Question

Solve using identities [0,2) cos2x=sinx

Answer 1

We have:

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

Thus:

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

Adding $2 \sin^2 x - 1$ gives:

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

Factoring gives:

$(2 \sin x + 1)(\sin x - 1) = 0$

Thus, we have:

$\sin x = \frac{-1}{2}$

or

$\sin x = 1$.

Note that because $2 \pi > 2$, there will be no solutions $y$ of the form $2n\pi + z$ for positive integer $n$ and solution $z$.  So we find the basic values of $x$ giving $\sin x \in (\frac{-1}{2}, 1)$.

For $\sin x = \frac{-1}{2}$ with positive $x$, $x > \pi > 2$ (this can be seen easily from the unit circle.  But $x < 2$, a contradiction.  So we examine the case in which $\sin x = 1$.

In this case, $x = \frac{\pi}{2}$.  So this is the only solution to this equation.

Substituting this back in, we have $\sin \frac{\pi}{2} = \cos \pi = 1$, as desired.  So this is a valid solution.

Thus, $x = \frac{\pi}{2}$.