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Chapter 5: Problem 9

Verify that the \(x\) -values are solutions of the equation. \(2 \sin ^{2} x-\sin x-1=0\) (a) \(x=\frac{\pi}{2}\) (b) \(x=\frac{7 \pi}{6}\)

Short Answer

Expert verified
Both \(x = \frac{\pi}{2}\) and \(x = \frac{7\pi}{6}\) are valid solutions to the equation \(2 \sin^2 x - \sin x - 1 = 0\).

Step by step solution

01

Substitute \(x = \frac{\pi}{2}\) into the equation

For part (a), substitute \(x = \frac{\pi}{2}\) into the equation \(2 \sin^2 x - \sin x - 1 = 0\). This gives us \(2 \sin^2 \frac{\pi}{2} - \sin \frac{\pi}{2} - 1 = 0\). Next, simplify this expression. We know that \(\sin \frac{\pi}{2} = 1\), so our equation becomes \(2(1)^2 - 1 - 1 = 0\). Simplifying further, we get \(2 - 1 - 1 = 0\), which simplifies to \(0 = 0\). Thus, \(x = \frac{\pi}{2}\) is indeed a solution to the equation.
02

Substitute \(x = \frac{7\pi}{6}\) into the equation

For part (b), substitute \(x = \frac{7\pi}{6}\) into the equation \(2 \sin^2 x - \sin x - 1 = 0\). This gives us \(2 \sin^2 \frac{7\pi}{6} - \sin \frac{7\pi}{6} - 1 = 0\). Let's simplify this expression. We know that \(\sin \frac{7\pi}{6} = -1/2\), so our equation becomes \(2(-1/2)^2 - (-1/2) - 1 = 0\). Simplifying further, we get \(2(1/4) + 1/2 - 1 = 0\), which simplifies to \(1/2 + 1/2 - 1 = 0\), thus again \(0 = 0\), confirming that \(x = \frac{7\pi}{6}\) is also a solution to the equation.

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