# Solve $\cos 2x - \sin x = 0$ for $0 \le x \le 360$

Solve $\cos 2x - \sin x$ for $0 \le x \le 360$

Using double angle formula:

$1 - 2\sin^2 x - \sin x = 0$

=> $2\sin^2 x + \sin x -1 = 0$

=> $(2\sin x - 1)(\sin x+ 1)$

=> $2\sin x = 1$

=> $2x = \arcsin(\frac{1}2)$

=> $x = 15^{\circ}$

$\sin x = -1$

$\sin$ is negative in 3rd and 4th quadrant

$x = 270^{\circ}$

My answers for $x$ are not correct

From $2 \sin x=1$, you should have $\sin x=0.5$. Sine is positive in the first two quadrants, you should obtain $30^{\circ}$ and $150^{\circ}$ as your solution as well.

• what about $\sin x = -1$? May 9, 2016 at 18:11
• I thought you have solved that? 270? May 9, 2016 at 18:13
• ok, I was not sure if that was correct May 9, 2016 at 18:14

Alternative route (avoids polynomial relations like double angle formula):

If $\cos\alpha=\cos\beta$ then $\alpha=\pm\beta+k\cdot360^{\circ}$ for $k\in\mathbb{Z}$.

We have $\cos2x=\sin x=\cos\left(x-90^{\circ}\right)$.

So $2x=\pm\left(x-90^{\circ}\right)+k\cdot360^{\circ}$.

Working out $2x=x-90^{\circ}+k\cdot360^{\circ}$ under condition $0^{\circ}\leq x\leq360^{\circ}$ leads to $x=270^{\circ}$.

Working out $2x=-x+90^{\circ}+k\cdot360^{\circ}$ under condition $0^{\circ}\leq x\leq360^{\circ}$ leads to $x\in\left\{ 30^{\circ},150^{\circ},270^{\circ}\right\}$.

we have $$\cos(2x)-\sin(x)=-2 \sin ^2\left(\frac{x}{2}+\frac{\pi }{4}\right) (2 \sin (x)-1)$$ and from here it is not difficult to solve the given equation