Chapter 7: Q. 91 (page 496)

In Problems 91-93, find the real zeros of each trigonometric function on the interval $0 \leq θ < 2 π$
$f (x) = \sin (2 x) - \sin x$

Short Answer

Expert verified

The solution set is $\{0, \frac{π}{3}, π, \frac{5 π}{3}\}$

Step by step solution

Given information

Given function $f (x) = \sin (2 x) - \sin x$

Substitute the values and take common

Substituting, we get

$f (x) = \sin (2 x) - \sin x f (x) = 2 \sin x \cos x - \sin x f (x) = \sin x (2 \cos x - 1) \sin x (2 \cos x - 1) = 0 \sin x = 0, 2 \cos x - 1 = 0 x = 0, π x = \frac{π}{3}, \frac{5 π}{3}$

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In Problems 91-93, find the real zeros of each trigonometric function on the interval $0 \leq θ < 2 π$
$f (x) = \sin (2 x) - \sin x$

Short Answer

Step by step solution

Given information

Substitute the values and take common

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In Problems 91-93, find the real zeros of each trigonometric function on the interval 0≤θ<2πf(x)=sin(2x)-sinx

Short Answer

Step by step solution

Given information

Substitute the values and take common

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Probability and Statistics

Logic and Functions

Decision Maths

Geometry

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

In Problems 91-93, find the real zeros of each trigonometric function on the interval $0 \leq θ < 2 π$
$f (x) = \sin (2 x) - \sin x$