# How do you find the general solutions for sinx = cos2x?

$x = {30}^{\setminus \circ} , {270}^{\setminus \circ}$ $\left[{0}^{0} \setminus \le q x \setminus \le q \setminus {360}^{0}\right]$

#### Explanation:

We know,

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

So, let's solve the equation now,

$\sin x = \cos 2 x = 1 - 2 {\sin}^{2} x$
$\setminus \rightarrow 2 {\sin}^{2} x + \sin x - 1 = 0$
$\setminus \rightarrow 2 {\sin}^{2} x + 2 \sin x - \sin x - 1 = 0$
$\setminus \rightarrow 2 \sin x \left(\sin x + 1\right) - 1 \left(\sin x + 1\right) = 0$
$\setminus \rightarrow \left(2 \sin x - 1\right) \left(\sin x + 1\right) = 0$

Now,

$2 \sin x - 1 = 0$
$\setminus \rightarrow \sin x = \setminus \frac{1}{2}$
\rightarrow x=sin^{-1}($\setminus \frac{1}{2}$)
$\setminus \rightarrow x = {30}^{\setminus \circ}$

And, $\sin x + 1 = 0$
$\setminus \rightarrow x = {\sin}^{- 1} \left(- 1\right)$$= {270}^{\setminus \circ}$

As we just need the general solutions, we should take only this two values as the general solutions .
Answer : ${30}^{0} , {270}^{0}$
That's it!

Sep 18, 2015

Solve sin x = cos 2x

#### Explanation:

Apply trig identity: $\cos 2 x = 1 - 2 {\sin}^{2} x$
$\sin x = 1 - 2 {\sin}^{2} x$. Solve the quadratic equation:
$2 {\sin}^{2} x + \sin x - 1 = 0$
Since (a - b + c = 0), use Shortcut. Two real roots: sin x = -1 and $\sin x = - \frac{c}{a} = \frac{1}{2}$.

$\sin x = \frac{1}{2}$ --> x = 30 deg and x = 150 deg $\left(\frac{\pi}{6} \mathmr{and} \frac{5 \pi}{6}\right)$
sin x = -1 --> x = 270 deg $\left(\frac{3 \pi}{2}\right)$
General solutions:
x = 30 + k360 deg
x = 150 + k360
x = 270 + k360.