Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 704: 105


The solution to the given equation is $\frac{\pi }{2}$.

Work Step by Step

We have to solve the equation on the interval $[0,2\pi )$: $\begin{align} & 2\text{si}{{\text{n}}^{2}}x=3-\text{sin}\ x \\ & 2\text{si}{{\text{n}}^{2}}x+\text{sin}\ x-3=0 \end{align}$ And simplify it as: $\begin{align} & 2{{\sin }^{2}}x-\sin x+3\sin x-3=0 \\ & 2\sin x\left( \sin x-1 \right)+3\left( \sin x-1 \right)=0 \\ & \left( \text{sin}x-1 \right)\left( 2\text{sin}x+3 \right)=0 \end{align}$ Each factor needs to be calculated as: $\begin{align} & \sin x-1=0 \\ & \sin x=0+1 \\ & \sin x=1 \end{align}$ Or $\begin{align} & 2\sin x+3=0 \\ & 2\sin x=0-3 \\ & \sin x=-\frac{3}{2} \end{align}$ So, in the quadrant graph, the value of sine is $1$ at $\frac{\pi }{2}$. This implies, $\begin{align} & sinx=sin\frac{\pi }{2} \\ & x=\frac{\pi }{2} \end{align}$ And in the quadrant graph, the value of sine counts to be $-\frac{3}{2}$. This value of sine lies between the values of $-1$ and $1$. Thus, the value of sine $-\frac{3}{2}$ will not be counted in the solution as it is less than the value of $-1$. These are the proposed solutions of the sine function. Thus, the actual solution in the interval $[0,2\pi )$ will be $\frac{\pi }{2}$.
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