Answered: For each of the following equations,…

Algebra and Trigonometry (MindTap Course List)

4th Edition

ISBN:9781305071742

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter7: Analytic Trigonometry

Section7.4: Basic Trigonometric Equations

Problem 2E: The basic equation sinx=2 has_________no/one/infinitely many solutions, whereas the basic equation...

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Question

For each of the following equations, solve for (a) all radian solutions and (b) x if
0 <=x , 2p. Give all answers as exact values in radians. Do not use a calculator.

(sin x - 1)(2 sin x - 1) = 0

Expert Solution

Step 1

Answer(a):

The given trigonometric equation is: $(\sin x - 1) (2 \sin x - 1) = 0$

That is,

Either $(\sin x - 1) = 0$ or $(2 \sin x - 1) = 0$

$\Rightarrow \sin x = 1$ or $\sin x = \frac{1}{2}$

$\Rightarrow \sin x = \sin \frac{π}{2}$ or $\sin x = \sin \frac{π}{6}$

The general solution of $\sin x = \sin α$ is given by:

$x = n π + {(- 1)}^{n} α$

Step 2

So, for $\sin x = \sin \frac{π}{2}$ , the general solution is:

$x = n π + {(- 1)}^{n} \frac{π}{2}$

And, for $\sin x = \sin \frac{π}{6}$ , the general solution is:

$x = n π + {(- 1)}^{n} \frac{π}{6}$

Therefore, the general solution of the given equation is:

$x = [n π + {(- 1)}^{n} \frac{π}{2}] \cup [n π + {(- 1)}^{n} \frac{π}{6}]$

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