Question Video: Solving Trigonometric Equations by Squaring | Nagwa Question Video: Solving Trigonometric Equations by Squaring | Nagwa

# Question Video: Solving Trigonometric Equations by Squaring Mathematics • First Year of Secondary School

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By first squaring both sides, or otherwise, solve the equation 4 sin 𝜃 − 4 cos 𝜃 = √3, where 0° < 𝜃 ⩽ 360°. Be careful to remove any extraneous solutions. Give your answers to two decimal places.

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### Video Transcript

By first squaring both sides, or otherwise, solve the equation four sin 𝜃 minus four cos 𝜃 is equal to the square root of three, where 𝜃 is greater than zero degrees and less than or equal to 360 degrees. Be careful to remove any extraneous solutions. Give your answers to two decimal places.

The question advises that we approach the problem by first squaring both sides of the equation. Doing so, we obtain four sin 𝜃 minus four cos 𝜃 all squared is equal to root three squared. Distributing the parentheses and then collecting like terms on the left-hand side gives us 16 sin squared 𝜃 minus 32 sin 𝜃 cos 𝜃 plus 16 cos squared 𝜃. On the right-hand side, root three squared is equal to three.

Next, we recall the Pythagorean identity, which states that sin squared 𝜃 plus cos squared 𝜃 is equal to one. Simplifying the left-hand side further gives us 16 multiplied by sin squared 𝜃 plus cos squared 𝜃 minus 32 sin 𝜃 cos 𝜃. Replacing sin squared 𝜃 plus cos squared 𝜃 with one gives us the equation 16 minus 32 sin 𝜃 cos 𝜃 equals three. Subtracting 16 from both sides of this equation gives us negative 32 sin 𝜃 cos 𝜃 is equal to negative 13. We can then divide through by negative 32 such that sin 𝜃 cos 𝜃 equals 13 over 32.

We now have two equations in the two variables sin 𝜃 and cos 𝜃. This means that the system of equations can be solved simultaneously. Adding four cos 𝜃 to both sides of our original equation, we have four sin 𝜃 is equal to root three plus four cos 𝜃. Dividing both sides of this equation by four, we have sin 𝜃 is equal to root three plus four cos 𝜃 all divided by four.

After clearing some space, we will now consider how we can solve these two simultaneous equations. We will begin by substituting the expression for sin 𝜃 in equation two into equation one. This gives us root three plus four cos 𝜃 over four multiplied by cos 𝜃 is equal to 13 over 32. We can simplify this equation by firstly distributing the parentheses. We can then multiply through by 32, giving us eight root three cos 𝜃 plus 32 cos squared 𝜃 equals 13. Finally, subtracting 13 from both sides of this equation, we have the quadratic equation in terms of cos 𝜃 as shown.

This can be solved using the quadratic formula, where 𝑎 is 32, 𝑏 is eight root three, and 𝑐 is negative 13. Substituting in these values and then simplifying gives us cos of 𝜃 is equal to negative three plus or minus the square root of 29 all divided by eight. Taking the inverse cosine of both sides with positive root 29 gives us 𝜃 is equal to 62.829 and so on. To two decimal places, this is equal to 62.83 degrees. Taking the inverse cosine of our equation with negative root 29 gives us 𝜃 is equal to 152.829 and so on. This rounds to 152.83 degrees to two decimal places.

We were asked to give all solutions that are greater than or equal to zero degrees and less than or equal to 360 degrees. We therefore need to consider the symmetry of the cosine function such that the cos of 𝜃 is equal to the cos of 360 degrees minus 𝜃. Subtracting each of our values from 360 degrees gives us further solutions 297.17 degrees and 207.17 degrees to two decimal places.

We have therefore found four possible solutions to the given equation. However, we were reminded in the question to remove any extraneous solutions, these extra solutions that were created when we squared our original equation. We need to substitute each of our four solutions into the initial equation to check they are valid.

The initial equation was four sin 𝜃 minus four cos 𝜃 equals root three. Substituting 𝜃 equals 62.83 into the left side of our equation gives us an answer of root three. This means that this is a valid solution. However, when we substitute 𝜃 is equal to 152.83 degrees into the left-hand side of our equation, we do not get root three. This means that this is not a valid solution. Repeating this process for 207.17 degrees and 297.17 degrees, we see that 207.17 is a valid solution, whereas the fourth answer of 297.17 is not. We can therefore conclude that there are two solutions that satisfy the equation in the given interval of 𝜃, which are 62.83 and 207.17 degrees.

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