The number of Solutions in [0,2pi] such that (Sin (2x))^4 = 1/8 is?

Oct 4, 2017

There are 4 solutions.

Explanation:

Right now, we have: ${\sin}^{4} \left(2 x\right) = \frac{1}{8}$

First, we must remove the ""^4 by taking the fourth root of each side:
$\sqrt[4]{{\sin}^{4} \left(2 x\right)} = \sin \left(2 x\right)$

$\sin \left(2 x\right) = \sqrt[4]{\frac{1}{8}} = \frac{\sqrt[4]{1}}{\sqrt[4]{8}} = \pm \frac{1}{1.682} = \pm 0.594530321$

Now, we must take the arcsin of each side:
$\arcsin \left(\sin \left(2 x\right)\right) = \arcsin \left(\pm 0.594530321\right) \equiv 2 x = {\sin}^{- 1} \left(\pm 0.594530321\right)$

Since our coefficient of $x$ is 2, our range must be doubled to $\left[0 , 4 \setminus \pi\right]$

${\sin}^{- 1} \left(0.594530321\right) = 2 x = 0.20266198 \setminus \pi \mathmr{and} 2.20266198 \setminus \pi$
${\sin}^{- 1} \left(- 0.594530321\right) = 2 x = 1.79733802 \pi \mathmr{and} 3.79733802 \pi \setminus \pi$

Therefore, $x = 0.10133099 \setminus \pi , 0.89866901 \setminus \pi , 1.89866901 \setminus \pi \mathmr{and} 1.10133099 \setminus \pi$, so there are 4 solutions.