   Chapter 8.2, Problem 61E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Solving a Trigonometric Equation In Exercises 61-70, solve the equation for θ . Assume 0 ≤ θ ≤ 2 π . For some of the equations, you should use the trigonometric identities listed in this section. Use a spreadsheet or a graphing utility to verify your results. See Example 7. 2 sin 2 θ = 1

To determine

The values of θ for the trigonometric equation 2sin2θ=1 and verify the result by graphing utility tool.

Explanation

Given Information:

The provided trigonometric equation is, 2sin2θ=1.

Consider the trigonometric equation, 2sin2θ=1.

Simplify the provided trigonometry equation,

2sin2θ=1sin2θ=12sinθ=±12

Since, the trigonometry function sine is positive in first and second quadrants, while it is negative in third and fourth quadrants.

Thus, there are four solutions for trigonometric equation 2sin2θ=1 for 0θ2π, which can be mathematically determined as:

sinθ=12sinθ=sinπ4θ=π4=0.7853981

sinθ=12sinθ=sin3π4θ=3π4=2.35619449

sinθ=12sinθ=sin5π4θ=5π4=3.926990817

sinθ=12sinθ=sin7π4θ=7π4=5.4977871

Therefore, the four values of θ for the trigonometric equation 2sin2θ=1 are θ=π4, θ=3π4, θ=5π4 and θ=7π4

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