# Find the x

#### mathwhat

$$\displaystyle 2\cos(2x)-2\sin(x)=0$$
another one:

$$\displaystyle 2\sin(x)+\sin(2x)=0$$

Thank you very much (Hi)

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#### mathwhat

Thank you very much but i did it exactly like you did and in the book the answers are different...
for 1:

$$\displaystyle \frac{\pi}{6}$$

$$\displaystyle \frac{5}{6}\pi$$

and for 2:
$$\displaystyle \pi$$

$$\displaystyle 2\pi$$

#### harish21

Thank you very much but i did it exactly like you did and in the book the answers are different...
for 1:

$$\displaystyle \frac{\pi}{6}$$

$$\displaystyle \frac{5}{6}\pi$$

and for 2:
$$\displaystyle \pi$$

$$\displaystyle 2\pi$$
The answers you have stated here are correct for the interval from $$\displaystyle 0$$ to $$\displaystyle 2\pi$$

#### mr fantastic

MHF Hall of Fame
Small error: $$\displaystyle \cos (2x) = 1 - 2\sin^2 (x)$$ NOT $$\displaystyle 2\sin^2 (x) - 1$$. So for #1 you will get pi/6 and 5pi/6 ....