Solving $\sin(x)=\cos(2x)$

Question

Please write a step by step solution and if possible an explanation. Thanks!

All I did so far is:

$\sin(x)=\cos(2x)$

$\sin(x) - \cos(2x) = 0$

Don't know where to go from here...

Answer 1

$$\cos x=\sin(90^\circ-x)$$

$$\sin y=\cos(?-y)$$

Now use $$\cos C=\cos D$$ or $$\sin C=\sin D$$ from http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

Answer 2

Starting from your last equation:

$\sin x - \sin (\frac{\pi}{2}-2x) = 0$

Now you know how to continue.

Answer 3

We can solve the trigonometric equation as also: $$\sin x=\cos 2x$$ $$ \Rightarrow ~ \sin x=1-2\sin^2 x$$ $$\Rightarrow ~ 2\sin^2 x + \sin x -1=0$$ Now try to transform the the above trigonometric equation in polynomial form as:

Let $t=\sin x$ then above final equation becomes $$2t^2+t-1=0.$$ Solving this we get that $$t=\frac{1}{2},-1.$$ Now we know that $t=\sin x$, when $t=\frac{1}{2}$, then $\sin x=\frac{1}{2}$, which implies that $x=\sin^{-1}(\frac{1}{2})=\pi/6$ and when $t=-1$, then $\sin x=-1$, which implies that $x=\sin^{-1}(-1)=-\pi/2$.