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Chapter 5: Problem 19

Solve the equation. $$2 \sin ^{2} 2 x=1$$

Short Answer

Expert verified
The solution set for the equation is \(x=\frac{\pi}{8}+n\pi, \frac{3\pi}{8}+n\pi\), where 'n' is an integer.

Step by step solution

01

Isolate the \(\sin ^{2} 2x\)

The first step involves isolating \(\sin ^{2} 2x\). To do this, the equation \(2 \sin ^{2} 2x = 1\) can be divided through by 2, resulting in \(\sin ^{2} 2x = 0.5\).
02

Apply the square root

Next, we must get \(\sin 2x\). This is achieved by applying the square root to both sides, remembering that the square root has two possible values, positive and negative. Thus, we get \(\sin 2x = \pm \sqrt{0.5} = \pm \frac{1}{\sqrt{2}}\) = \(\pm \frac{\sqrt{2}}{2}\).
03

Use inverse sine function

In this step, take the inverse sine (\(arcsin\)) of both sides to determine the value of \(2x\). However, we remember that \(\sin(\pi -y) = \sin y\), so applying that here, we get \(2x = arcsin(\pm\frac{\sqrt{2}}{2}) = \frac{\pi}{4}, \frac{3\pi}{4}\).
04

Solve for x

Finally, we solve for 'x' by dividing through by 2. This yields \(x=\frac{\pi}{8}, \frac{3\pi}{8}\). However, since sine is a periodic function with a period of 2\(\pi\), any multiples of the period can be added to these solutions. So, the complete solution is \(x=\frac{\pi}{8}+n\pi, \frac{3\pi}{8}+n\pi\), where 'n' can be any integer.

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