Solve $7\sin^2(x) - 9\cos(2x) = 0$

Question

I need to solve for x in the polynomial

$$7\sin^2(x) - 9\cos(2x) = 0$$

I have tried approaching the problem in multiple ways. I am only looking for some hints, not the actual answer. Thanks :D

Answer 1

Two possibilities of solving this one.

1. Using $\cos(2x) = \cos^2(x)-\sin^2 x = (1-\sin^2 x) - \sin^2 x$ will give you an equation with only $\sin^2 x$ to solve
2. Using $\sin^2 x = \frac{1-\cos(2x)}{2}$ will give you an equation with only $\cos(2x)$.

Answer 2

There are two ways to solve this all leading to the same answer:

1) 7sin^2(x)−9cos(2x)=0

using $cos(2x) = $cos^2x - sin^2x$

=> $7sin^2(x)−9(cos^2x - sin^2x)=0$ {as cos(2x) = $cos^2x - sin^2x$}

=> $7sin^2(x)−9(cos^2x - sin^2x)=0$

=> $7sin^2(x)−9(1 - sin^2x - sin^2x)=0$ {as $cos^2x = 1 - sin^2x$}

=> $7sin^2(x)−9(1 - 2sin^2x)=0$

=> $7sin^2(x)−9 + 18sin^2x)=0$

=> $25sin^2x=9$

=> $sinx=(3/5)$ or $sinx= -(3/5)$

=> x = 0.643501109 rad or x = 36.8698976 degrees

2) 7sin^2(x)−9cos(2x)=0

 using $sin^2(x)=(1−cos(2x))/2$

=> $7(1−cos(2x))/2−9cos(2x)=0$

=> $7−7cos(2x)−18cos(2x)=0$

=> $7=25cos(2x)$

=> $cos(2x) = 7/25$

=> $2x= arccos(7/25)$ {which is 1.28700222 rad or 73.7397953 degrees}

=> $x=$ 0.64350111 rad or x = 36.86989765 degrees

Ref : http://en.wikipedia.org/wiki/List_of_trigonometric_identities