Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 16

Answer

There are only 1 value of $x$ satisfying the equation, which is $$x=\frac{\pi}{2}$$

Work Step by Step

$$\sin x+2=3$$ $$\sin x=1$$ In the interval $[0,2\pi)$, there are only 1 value of $x$ satisfying the equation, which is $$x=\frac{\pi}{2}$$ *NOTES: For $\sin x=a$, except when $\sin x=0$, the set of values of satisfying $x$ is $x=u+2\pi$, which means for $\sin x$ to reach a specific amount again, the cycle has to go through $2\pi$, or in fact a full circle.
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