Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.

In this lesson, we will learn how to solve trigonometric equations by squaring both sides.

Q1:

Consider the equation s i n c o s π + π = β 2 , where 0 < π β€ 3 6 0 β β . Call this Equation A.

Create Equation B by squaring both sides of Equation A. Use the fact that s i n c o s 2 2 π + π = 1 to simplify Equation B.

Now, use a double angle formula to further simplify Equation B.

The solutions to Equation A are a subset of the solutions of Equation B. Using this, solve Equation A over the specified range.

Q2:

Solve β 2 π + β 3 π = 2 s i n c o s , where 0 < π β€ 2 π . Give your answer in radians to three significant figures.

Q3:

If π β [ 0 , 1 8 0 [ β β and s i n c o s π + π = 1 , find the possible values of π .

Q4:

By first squaring both sides, or otherwise, solve the equation 4 π β 4 π = β 3 s i n c o s , where 0 < π β€ 3 6 0 . Be careful to remove any extraneous solutions. Give your answers to two decimal places.

Q5:

Find the set of values satisfying 9 7 π + 6 0 π = 0 s i n c o s where 0 < π < 3 6 0 β β . Give the answers to the nearest second.

Q6:

Find the set of values satisfying s i n c o s π β π = 0 where 0 < π < 3 6 0 β β . Give the answers to the nearest second.

Q7:

If π β ( 1 8 0 , 3 6 0 ) β β and s i n c o s π + π = β 1 , find the value of π .

Donβt have an account? Sign Up