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Chapter 5: Problem 75

Find all solutions of the equation in the interval \([0,2 \pi)\). $$\sin (x+\pi)-\sin x+1=0$$

Short Answer

Expert verified
The solutions to the equation in the interval [0, 2蟺) are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

Step by step solution

01

Utilize Trigonometric Identity

Use the identity \(\sin (x + \pi) = -\sin(x)\) to simplify the expression. The equation becomes: \(-\sin x - \sin x + 1 = 0.\)
02

Simplify the Equation

Combine the like terms to simplify this to: \(-2\sin x + 1 = 0.\)
03

Solve for \(x\)

Solve for \(x\) by first subtracting 1 from both sides, then dividing by -2. This gives: \(\sin x = \frac{1}{2}\).
04

Determine the Values of \(x\)

For \(x\) to satisfy the equation in the interval \([0,2 \pi)\), we know from the unit circle that \(\sin x = \frac{1}{2}\) when \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

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