Lesson Explainer: Solving Trigonometric Equations with the Double-Angle Identity Mathematics

In this explainer, we will learn how to solve trigonometric equations using the double-angle identity.

Trigonometric equations have several real-world applications in various fields, such as physics, engineering, architecture, robotics, music theory, and navigation, to name a few. In physics, they can be used in projectile motion, modeling the mechanics of electromagnetic waves, analyzing alternating and direct currents, and finding the trajectory of a mass around a massive body under the force of gravity.

Let’s begin by recalling the trigonometric functions, whose double-angle identities we will examine in this explainer. Consider a right triangle.

The trigonometric functions can be expressed in terms of the ratio of the sides of the triangle as sinOHcosAHtanOAπœƒ=,πœƒ=,πœƒ=.

These functions satisfy the following trigonometric identity: tansincosπœƒ=πœƒπœƒ.

We note that these trigonometric ratios are defined for acute angles 0<πœƒ<90∘∘, and the trigonometric functions for all values of πœƒ are defined on the unit circle using right triangle trigonometry.

Suppose that a point moves along the unit circle in the counterclockwise direction. At a particular position (π‘₯,𝑦) on the unit circle with angle πœƒ, the sine function is defined as 𝑦=πœƒsin and the cosine function as π‘₯=πœƒcos, as shown in the diagram above. In other words, the trigonometric functions are defined using the coordinates of the point of intersection of the unit circle with the terminal side of πœƒ in standard position.

The domain is the set of possible inputs and the range is the set of possible outputs, given its domain. For the trigonometric functions, these are given by the following.

DomainRange
sinπœƒβ„[βˆ’1,1]
cosπœƒβ„[βˆ’1,1]
tanπœƒβ„βˆ’ο“πœ‹2+π‘›πœ‹,π‘›βˆˆβ„€οŸβ„

Since the tangent function is defined as the ratio of the sine and cosine functions, it is undefined when cosπœƒ, the denominator, is zero. In other words, the tangent function has to exclude values of πœƒ where cosπœƒ=0 in order to be well defined. This is why the domain of the tangent function is β„βˆ’ο“πœ‹2+π‘›πœ‹,π‘›βˆˆβ„€οŸ, which just means we subtract the values of πœƒ where cosπœƒ=0 from the set of real numbers in order to exclude this from the input.

The trigonometric functions are periodic, which means if we add an integer multiple of 2πœ‹, in radians, or 360∘ to the angle πœƒ, the value of the function stays the same: sinsincoscostantan(360+πœƒ)=πœƒ,(360+πœƒ)=πœƒ,(360+πœƒ)=πœƒ.∘∘∘

We can see these directly from the unit circle definition of the trigonometric functions. In fact, the tangent function is periodic by πœ‹, in radians, or 180∘ since we have tantan(180+πœƒ)=πœƒ.∘

This fact will be important for finding general solutions for the trigonometric functions. The domains of the trigonometric functions have to be restricted to a particular subset, known as the principle branch, in order to have inverse functions.

The inverse trigonometric functions denoted by sin, cos, and tan are the inverse functions of the trigonometric functions sin, cos, and tan. This means they work in reverse or β€œgo backward” from the usual trigonometric functions. They are defined by 𝑦=π‘₯⟺π‘₯=𝑦,𝑦=π‘₯⟺π‘₯=𝑦,𝑦=π‘₯⟺π‘₯=𝑦.sinsincoscostantan

These can also be written as arcsinπ‘₯, arccosπ‘₯, and arctanπ‘₯. The domain and range for the inverse trigonometric functions are given by the following.

DomainRange
sinοŠ±οŠ§πœƒ[βˆ’1,1]ο“βˆ’πœ‹2,πœ‹2
cosοŠ±οŠ§πœƒ[βˆ’1,1][0,πœ‹]
tanοŠ±οŠ§πœƒβ„οŸβˆ’πœ‹2,πœ‹2

The ranges for the inverse trigonometric functions in general apply only when the trigonometric functions are restricted to the principal branch. This is to ensure that the trigonometric functions are one-to-one functions, so that the inverse trigonometric functions evaluate to a single value, known as the principal value.

For example, if we have a particular trigonometric equation, such as sinπœƒ=𝑦, we can find the solutions in the range πœƒβˆˆο“βˆ’πœ‹2,πœ‹2 by applying the inverse trigonometric equations: πœƒ=(𝑦).sin

However, if we want to determine all the possible solutions, we need the general solutions given in terms of an integer π‘›βˆˆβ„€, which we can obtain from the CAST diagram and the periodicity of the trigonometric functions.

Let’s recall the CAST diagram.

The Cast Diagram

  • In the first quadrant, all trigonometric functions are positive.
  • In the second quadrant, the sine function is positive.
  • In the third quadrant, the tangent function is positive.
  • In the fourth quadrant, the cosine function is positive.

Let’s recall how we can find the solutions to trigonometric equations.

Solutions to Trigonometric Equations

The CAST diagram helps us to remember the signs of the trigonometric functions for each quadrant.

In particular, the CAST diagram tells us that solutions to the trigonometric equations are given by the following.

  • If sinπœƒ=π‘₯ and βˆ’1≀π‘₯≀1, πœƒ=π‘₯πœƒ=ο€Ί180βˆ’π‘₯,sinsin∘ for πœƒβˆˆ[βˆ’90,270]∘∘, or, in radians, πœƒ=π‘₯πœƒ=ο€Ίπœ‹βˆ’π‘₯,sinsin for πœƒβˆˆο”βˆ’πœ‹2,3πœ‹2.
  • If cosπœƒ=π‘₯ and βˆ’1≀π‘₯≀1, then we can express the angle πœƒ in terms of the inverse cosine function in degrees as πœƒ=π‘₯πœƒ=ο€Ή360βˆ’π‘₯,coscos∘ for πœƒβˆˆ[0,360]∘, or, in radians, πœƒ=π‘₯πœƒ=ο€Ή2πœ‹βˆ’π‘₯,coscos for πœƒβˆˆ[0,2πœ‹].
  • If tanπœƒ=π‘₯, then we can express the angle πœƒ in terms of the inverse tangent function in degrees as πœƒ=π‘₯πœƒ=ο€Ή180+π‘₯,tantan∘ for πœƒβˆˆ]βˆ’90,90[βˆͺ]90,270[∘∘∘∘, or, in radians, πœƒ=π‘₯πœƒ=ο€Ήπœ‹+π‘₯,tantan for πœƒβˆˆοŸβˆ’πœ‹2,πœ‹2βˆͺο πœ‹2,3πœ‹2.

The ranges given for πœƒ follow from the ranges of the inverse trigonometric functions.

We can also see this from the unit circle as shown.

The general solutions to the trigonometric equations can be found from the solutions we obtain from the CAST diagram or inverse trigonometric functions, πœƒ, by adding a integer multiple of 360∘ or 2πœ‹, in radians. We do this for all the solutions we obtain, since the trigonometric functions are periodic. Thus, the general solution, Μ‚πœƒ, for π‘›βˆˆβ„€, is Μ‚πœƒ=πœƒ+360π‘›βˆ˜ in degrees and Μ‚πœƒ=πœƒ+2πœ‹π‘› in radians.

When solving trigonometric equations, we are usually given a particular range for the angle πœƒ to determine the solutions, which means we may only need to consider a few values of 𝑛, where appropriate. A solution set is the set of values that contains solutions to the trigonometric equation in the required range.

Now, let’s recall the addition identities for the sine, cosine, and tangent functions: sinsincoscossincoscoscossinsintantantantantan(πœƒΒ±πœƒ)=πœƒπœƒΒ±πœƒπœƒ,(πœƒΒ±πœƒ)=πœƒπœƒβˆ“πœƒπœƒ,(πœƒΒ±πœƒ)=πœƒΒ±πœƒ1βˆ’πœƒπœƒ.

We will use these addition identities to derive the double-angle identities.

By substituting πœƒ=πœƒ=πœƒοŠ¨οŠ§ in the addition identities, we obtain the double-angle identities for the trigonometric functions.

Definition: Trigonometric Double-Angle Identities

The trigonometric double-angle identities are sinsincoscoscossintantantan2πœƒ=2πœƒπœƒ,2πœƒ=πœƒβˆ’πœƒ,2πœƒ=2πœƒ1βˆ’πœƒ.

Let’s consider an example that demonstrates how we can use the sine double-angle identity to solve a particular trigonometric equation in a specified range.

Example 1: Solving an Equation in a Specified Range Using the Double-Angle Identities

Find the set of possible solutions of 2πœƒπœƒ=0sincos given πœƒβˆˆ[0,360[∘∘.

Answer

In this example, we are going to solve a trigonometric equation in a particular range using the double-angle identities.

The double-angle formula for sine is given by sinsincos2πœƒ=2πœƒπœƒ.

Therefore, 2πœƒπœƒ=0sincos is equivalent to sin2πœƒ=0.

The general solution, for π‘›βˆˆβ„€ (using the CAST diagram), to this equation can be found as 2πœƒ=0+360π‘›βˆ˜βˆ˜ or 2πœƒ=(180βˆ’0)+360𝑛,∘∘∘ which is equivalent to πœƒ=180π‘›βˆ˜ or πœƒ=90+180𝑛,∘∘ for π‘›βˆˆβ„€. The first expression gives us πœƒ=0,180∘∘ and the second expression πœƒ=90,270∘∘, for 𝑛=0,1. For other integers 𝑛, we would obtain angles outside the required range.

Thus, given πœƒβˆˆ[0,360[∘∘, the possible solutions are {0,90,180,270}.∘∘∘∘

The double-angle identity can also be used to solve trigonometric equations of the form that we saw earlier, π‘Žπœƒ+π‘πœƒ=𝑐,sincos by squaring both sides and using the Pythagorean identity. Now, let’s consider an example where we demonstrate this to find the solutions of a trigonometric equation in this form.

Example 2: Solving Trigonometric Equations Involving Special Angles

If πœƒβˆˆ]180,360[∘∘ and sincosπœƒ+πœƒ=βˆ’1, find the value of πœƒ.

Answer

In this example, we are going to solve a trigonometric equation in a particular range using the double-angle identities.

In order to solve sincosπœƒ+πœƒ=βˆ’1, we begin by squaring both sides and distributing: (πœƒ+πœƒ)=(βˆ’1)πœƒ+πœƒ+2πœƒπœƒ=1.sincossincossincos

On applying the Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒ=1 and the double-angle identity sinsincos2πœƒ=2πœƒπœƒ, we have 1+2πœƒ=12πœƒ=0.sinsin

The general solution, for π‘›βˆˆβ„€ (using the CAST diagram), to this equation can be found as 2πœƒ=(0)+360𝑛=360𝑛sin∘∘ or 2πœƒ=ο€Ί180βˆ’(0)+360𝑛=180+360𝑛,∘∘∘∘sin which is equivalent to πœƒ=180π‘›βˆ˜ or πœƒ=90+180𝑛,∘∘ for π‘›βˆˆβ„€. The second expression gives πœƒ=270∘, for 𝑛=1. For other integers 𝑛, we would obtain angles outside the required range.

Thus, given πœƒβˆˆ]180,360[∘∘, the only possible solution is πœƒ=270.∘

Let’s see how we can use the double-angle identities to solve other trigonometric equations in a particular range. As an example, suppose we want to find all the solutions within the range πœƒβˆˆ[0,720]∘∘ to the trigonometric equation 9πœƒβˆ’4=2πœƒ.sincos

On applying the double-angle formula for cosine and using the Pythagorean identity, this can be written as 9πœƒβˆ’4=2πœƒ=πœƒβˆ’πœƒ=ο€Ί1βˆ’πœƒο†βˆ’πœƒ=1βˆ’2πœƒ2πœƒ+9πœƒβˆ’5=0.sincoscossinsinsinsinsinsin

If we let 𝑦=πœƒsin, this is equivalent to solving the quadratic equation 2𝑦+9π‘¦βˆ’5=0.

We can solve this using the quadratic formula or factorization to give 2𝑦+9π‘¦βˆ’5=(2π‘¦βˆ’1)(𝑦+5)=0.

Thus, the solutions are 𝑦=12 and 𝑦=βˆ’5. We can ignore the second solution since for 𝑦=πœƒsin, we have βˆ’1≀𝑦≀1. Thus, we only consider the solutions with 𝑦=12 or sinπœƒ=12, for πœƒβˆˆ[0,720]∘∘. The acute solution is given by πœƒ=ο€Ό12=30.sin∘

General solutions can be found using the CAST diagram and the periodicity of the sine function, for π‘›βˆˆβ„€, as πœƒ=ο€Ό12+360𝑛=30+360𝑛sin∘∘∘ and πœƒ=ο€Ό180βˆ’ο€Ό12+360𝑛=(180βˆ’30)+360𝑛=150+360𝑛.∘∘∘∘∘∘∘sin

Now, we can substitute particular integer values of 𝑛 in order to find all the solutions within the required range. In particular, for 𝑛=0 and 𝑛=1, we obtain the solutions πœƒ=30,390∘∘ and πœƒ=150,510∘∘ from the first and second expressions for the general solution respectively. For other integers 𝑛, we would obtain angles outside the range [0,720]∘∘.

To summarize, the solutions to 9πœƒβˆ’4=2πœƒsincos, in degrees, for πœƒβˆˆ[0,720]∘∘, are πœƒ=30,150,390,510.∘∘∘∘

Now, let’s look at a few more examples in order to practice and deepen our understanding of solving trigonometric equations using the double-angle identities.

In the next example, we will use the double-angle identity for sine to find the solutions, in degrees.

Example 3: Solving a Trigonometric Equation Using Double-Angle Identities

Find the set of solutions in the range 0<π‘₯<180∘∘ for the equation (π‘₯+π‘₯)=22π‘₯sincossin.

Answer

In this example, we are going to solve a trigonometric equation in a particular range using the double-angle identities.

On distributing the parentheses on the left-hand side of the given trigonometric equation and applying the Pythagorean identity sincosπ‘₯+π‘₯=1, (π‘₯+π‘₯)=π‘₯+2π‘₯π‘₯+π‘₯=1+2π‘₯π‘₯.sincossinsincoscossincos

The double-angle identity for sine is given by sinsincos2π‘₯=2π‘₯π‘₯.

On substituting this, we have (π‘₯+π‘₯)=1+2π‘₯π‘₯=1+2π‘₯,sincossincossin and thus the given trigonometric equation (π‘₯+π‘₯)=22π‘₯sincossin is equivalent to 1+2π‘₯=22π‘₯22π‘₯βˆ’2π‘₯βˆ’1=0.sinsinsinsin

If we let 𝑦=2π‘₯sin, we have 2π‘¦βˆ’π‘¦βˆ’1=0.

The solutions are 𝑦=1 and 𝑦=βˆ’12. For 𝑦=1, we have sin2π‘₯=1, which has the general solution, for π‘›βˆˆβ„€ (using the CAST diagram), 2π‘₯=(1)+360𝑛=90+360𝑛π‘₯=45+180𝑛sin∘∘∘∘∘ and 2π‘₯=ο€Ί180βˆ’(1)+360𝑛=(180βˆ’90)+360𝑛=90+360𝑛π‘₯=45+180𝑛.∘∘∘∘∘∘∘∘∘sin

These two expressions are equivalent. For 𝑦=βˆ’12, we have sin2π‘₯=βˆ’12, which has the general solution 2π‘₯=ο€Όβˆ’12+360𝑛=βˆ’30+360𝑛π‘₯=βˆ’15+180𝑛sin∘∘∘∘∘ and 2π‘₯=ο€Ό180βˆ’ο€Όβˆ’12+360𝑛=(180+30)+360𝑛=210+360𝑛π‘₯=105+180𝑛.∘∘∘∘∘∘∘∘∘sin

To summarize, the general solutions, for π‘›βˆˆβ„€, are π‘₯=45+180𝑛,π‘₯=βˆ’15+180𝑛,π‘₯=105+180𝑛.∘∘∘∘∘∘

For 𝑛=0, we obtain the solutions π‘₯=45∘ and π‘₯=105∘, from the first and third expression respectively, and for 𝑛=1, we obtain π‘₯=165∘ from the second solution. For other integers 𝑛, we would obtain angles outside the required range.

Thus, given 0≀π‘₯<180∘∘, the solutions are {45,105,165}.∘∘∘

Now, let’s look at an example where we will use the double-angle identity for cosine to find the solutions, in degrees, to a trigonometric equation. This time we also have to consider a quadratic equation and the range for the cosine function.

Example 4: Using Double-Angle Identities to Solve a Trigonometric Equation

Find the solution set for π‘₯ given coscos2π‘₯+13√3π‘₯=βˆ’19, where π‘₯∈]0,360[∘.

Answer

In this example, we are going to solve a trigonometric equation in a particular range using the double-angle identities.

The double-angle formula for cosine is given by coscossin2π‘₯=π‘₯βˆ’π‘₯.

On applying the Pythagorean identity, we can rewrite this as coscoscoscos2π‘₯=π‘₯βˆ’ο€Ή1βˆ’π‘₯=2π‘₯βˆ’1.

On applying this to the left-hand side of the given trigonometric equation, we obtain coscoscoscos2π‘₯+13√3π‘₯=2π‘₯βˆ’1+13√3π‘₯.

Thus, the given trigonometric equation coscos2π‘₯+13√3π‘₯=βˆ’19 can be rewritten as 2π‘₯βˆ’1+13√3π‘₯=βˆ’192π‘₯+13√3π‘₯+18=0.coscoscoscos

If we let 𝑦=π‘₯cos, this can be written as a quadratic equation: 2𝑦+13√3𝑦+18=0.

We can find the solution to this quadratic equation using the quadratic formula to obtain 𝑦=βˆ’13√3Β±ο„žο€»13√3ο‡βˆ’4Γ—2Γ—182Γ—2=βˆ’13√3±√3634.

This gives 𝑦=βˆ’6√3 and 𝑦=βˆ’βˆš32, but since 𝑦=π‘₯cos and we have βˆ’1≀𝑦≀1, then we can ignore the first solution and have to solve cosπ‘₯=βˆ’βˆš32.

The general solution, for π‘›βˆˆβ„€ (using the CAST diagram), can be written as π‘₯=ο€Ώβˆ’βˆš32+360𝑛=150+360𝑛cos∘∘∘ and π‘₯=ο€Ώ360βˆ’ο€Ώβˆ’βˆš32+360𝑛=(360βˆ’150)+360𝑛=210+360𝑛.∘∘∘∘∘∘∘cos

The first expression gives π‘₯=150∘ and the second expression π‘₯=210∘, for 𝑛=0. For other integers 𝑛, we would obtain angles outside the required range.

Thus, given π‘₯∈]0,360[∘∘, the possible solutions are {150,210}.∘∘

In the next example, we will see how we can use either the double-angle sine or cosine identity to solve a trigonometric equation, as it can be expressed in terms of both after some manipulation.

Example 5: Solving a Trigonometric Equation Using Double-Angle Identities

Find the set of possible values of π‘₯ that satisfy 1√π‘₯βˆ’π‘₯=2coscosοŠͺ, where 0<π‘₯<360∘∘.

Answer

In this example, we are going to solve a trigonometric equation in a particular range using the double-angle identities.

Recall that the double-angle identity for cosine is sinsincos2π‘₯=2π‘₯π‘₯.

Now, to solve the given trigonometric equation, we note that, using the Pythagorean identity sincosπ‘₯=1βˆ’π‘₯, the denominator on the left-hand side can be written as √π‘₯βˆ’π‘₯=√π‘₯(1βˆ’π‘₯)=√π‘₯π‘₯=π‘₯π‘₯.coscoscoscoscossincossinοŠͺ

Thus, using the double-angle identity, the given trigonometric equation becomes 1π‘₯π‘₯=22π‘₯π‘₯=12π‘₯=1,cossinsincossin which has the general solution, for π‘›βˆˆβ„€ (using the CAST diagram), 2π‘₯=(1)+360𝑛=90+360𝑛π‘₯=45+180𝑛sin∘∘∘∘∘ and 2π‘₯=ο€Ί180βˆ’(1)+360𝑛=(180βˆ’90)+360𝑛=90+360𝑛π‘₯=45+180𝑛.∘∘∘∘∘∘∘∘∘sin

These two expressions are equivalent. This gives us πœƒ=45,135,225,315∘∘∘∘ for 𝑛=0,1,2, and 3. For other integers 𝑛, we would obtain angles outside the required range.

Thus, given 0<π‘₯<360∘∘, the possible solutions are {45,135,225,315}.∘∘∘∘

Some trigonometric equations may require the use of the half-angle identities for the trigonometric functions, which follow directly from the double-angle identities.

The half-angle identities for the trigonometric functions are given by the following.

Definition: Trigonometric Half-Angle Identities

The trigonometric half-angle identities are sincoscoscostancossinο€½πœƒ2=Β±ο„ž1βˆ’πœƒ2,ο€½πœƒ2=Β±ο„ž1+πœƒ2,ο€½πœƒ2=1βˆ’πœƒπœƒ.

These can be shown from the double-angle identities. For example, if we consider the double-angle identity for cosine, coscossinsin(2π‘₯)=π‘₯βˆ’π‘₯=1βˆ’2π‘₯, we rearrange this to make sinπ‘₯ the subject of the equation: 2π‘₯=1βˆ’2π‘₯π‘₯=1βˆ’2π‘₯2π‘₯=Β±ο„ž1βˆ’2π‘₯2.sincossincossincos

Now, if we let π‘₯=πœƒ2, this can be written as sincosο€½πœƒ2=Β±ο„ž1βˆ’πœƒ2, which is the half-angle identity for sine. The other half-angle identities can be found in a similar way.

Now, let’s look at an example where we use the half-angle identity for cosine to solve a trigonometric equation, in radians.

Example 6: Solving Trigonometric Equations Involving Half Angles

Using the half-angle formula sincosο€»π‘₯2=ο„ž1βˆ’π‘₯2, or otherwise, solve the equation sincosο€»π‘₯2+π‘₯=1, where 0≀π‘₯<2πœ‹.

Answer

In this example, we will solve a trigonometric equation in a particular range using the half-angle identity for cosine.

If we substitute the half-angle formula, the equation can be rewritten as ο„ž1βˆ’π‘₯2+π‘₯=1ο„ž1βˆ’π‘₯2=1βˆ’π‘₯.coscoscoscos

On squaring both sides, we have 1βˆ’π‘₯2=(1βˆ’π‘₯)1βˆ’π‘₯=2(1βˆ’π‘₯)1βˆ’π‘₯=2βˆ’4π‘₯+2π‘₯2π‘₯βˆ’3π‘₯+1=0.coscoscoscoscoscoscoscoscos

If we let 𝑦=π‘₯cos, we have to solve 2π‘¦βˆ’3𝑦+1=0.

The solutions are 𝑦=1 and 𝑦=12. For 𝑦=1, we have cosπ‘₯=1, which has the general solution, for π‘›βˆˆβ„€ (using the CAST diagram), π‘₯=(1)+2πœ‹π‘›=0+2πœ‹π‘›=2πœ‹π‘›cos and π‘₯=ο€Ή2πœ‹βˆ’(1)+2πœ‹π‘›=(2πœ‹βˆ’0)+2πœ‹π‘›=2πœ‹+2πœ‹π‘›.cos

We note that the second expression is equivalent to the first; the general solution is an integer multiple of 2πœ‹. For 𝑦=12, we have cosπ‘₯=12, which has the general solution, for π‘›βˆˆβ„€, π‘₯=ο€Ό12+2πœ‹π‘›=πœ‹3+2πœ‹π‘›cos and π‘₯=ο€Ό2πœ‹βˆ’ο€Ό12+2πœ‹π‘›=ο€»2πœ‹βˆ’πœ‹3+2πœ‹π‘›=5πœ‹3+2πœ‹π‘›.cos

To summarize, the general solutions, for π‘›βˆˆβ„€, are π‘₯=2πœ‹π‘›,π‘₯=πœ‹3+2πœ‹π‘›,π‘₯=5πœ‹3+2πœ‹π‘›.

For 𝑛=0, we obtain the solutions π‘₯=0, π‘₯=πœ‹3, and π‘₯=5πœ‹3 from the first, second, and third expressions respectively. For other integers 𝑛, we would obtain angles outside the required range.

Thus, given 0≀π‘₯<2πœ‹, the solutions are π‘₯∈0,13πœ‹,53πœ‹οΈ.

Finally, we will look at an example where we use the half-angle identity for the tangent function to solve a trigonometric equation, in radians.

Example 7: Solving Trigonometric Equations Involving Half Angles

Solve tansinο€»π‘₯2=π‘₯, where 0≀π‘₯<2πœ‹.

Answer

In this example, we are going to solve a trigonometric equation in a particular range using the half-angle identities.

The half-angle identity for the tangent function is given by tancossinο€»π‘₯2=1βˆ’π‘₯π‘₯.

Thus, we have to solve 1βˆ’π‘₯π‘₯=π‘₯1βˆ’π‘₯=π‘₯.cossinsincossin

On applying the Pythagorean identity, we have 1βˆ’π‘₯=1βˆ’π‘₯π‘₯βˆ’π‘₯=0.coscoscoscos

If we let 𝑦=π‘₯cos, π‘¦βˆ’π‘¦=0𝑦(π‘¦βˆ’1)=0.

The solutions to this quadratic equation are 𝑦=0 and 𝑦=1. For 𝑦=0, we have cosπ‘₯=0, which has the general solution, for π‘›βˆˆβ„€ (using the CAST diagram), π‘₯=(0)+2πœ‹π‘›=πœ‹2+2πœ‹π‘›cos and π‘₯=ο€Ή2πœ‹βˆ’(0)+2πœ‹π‘›=ο€»2πœ‹βˆ’πœ‹2+2πœ‹π‘›=3πœ‹2+2πœ‹π‘›.cos

For 𝑦=1, we have cosπ‘₯=1, which has the general solution π‘₯=(1)+2πœ‹π‘›=0+2πœ‹π‘›=2πœ‹π‘›cos and π‘₯=ο€Ή2πœ‹βˆ’(1)+2πœ‹π‘›=(2πœ‹βˆ’0)+2πœ‹π‘›=2πœ‹+2πœ‹π‘›.cos

We note that the second expression is equivalent to the first, that the general solution is an integer multiple of 2πœ‹. To summarize, the general solutions, for π‘›βˆˆβ„€, are π‘₯=πœ‹2+2πœ‹π‘›,π‘₯=3πœ‹2+2πœ‹π‘›,π‘₯=2πœ‹π‘›.

For 𝑛=0, we obtain the solutions π‘₯=πœ‹2, π‘₯=3πœ‹2, and π‘₯=0 from the first, second, and third expressions respectively. For other integers 𝑛, we would obtain angles outside the required range.

Thus, given 0≀π‘₯<2πœ‹, the solutions are π‘₯∈0,πœ‹2,3πœ‹2.

Let us finish by recapping a few important key points from this explainer.

Key Points

  • We can solve trigonometric equations using the double-angle or half-angle identities.
  • The trigonometric double-angle identities are sinsincoscoscossintantantan2πœƒ=2πœƒπœƒ,2πœƒ=πœƒβˆ’πœƒ,2πœƒ=2πœƒ1βˆ’πœƒ.
  • The trigonometric half-angle identities are sincoscoscostancossinο€½πœƒ2=Β±ο„ž1βˆ’πœƒ2,ο€½πœƒ2=Β±ο„ž1+πœƒ2,ο€½πœƒ2=1βˆ’πœƒπœƒ. These follow directly from the double-angle identities.
  • After finding the principal value solution, in degrees or radians, we can find the general solution for the trigonometric functions, for π‘›βˆˆβ„€, using the CAST diagram and the periodicity of the trigonometric functions.
  • We are usually given a particular range for the angle πœƒ to determine the solutions, which means we only consider particular integers 𝑛 to find the possible solutions.

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