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Chapter 7: Q. 75 (page 496)

In Problems $69 - 78$ , solve each equation on the interval $0 \leq θ < 2 π$
$3 - \sin θ = \cos (2 θ)$

Short Answer

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Step by step solution

01

Given information

Given function $3 - \sin θ = \cos (2 θ)$

02

Converting to same form

Converting, we get

$3 - \sin θ = 1 - 2 \sin^{2} θ 3 - \sin θ + 2 \sin^{2} θ - 1 = 1 - 2 \sin^{2} θ + 2 \sin^{2} θ - 1 2 \sin^{2} θ - \sin θ + 2 = 0$

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Most popular questions from this chapter

Show that $\cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = 0$

$\cos (- θ) - \cos (θ)$

Express the product $\sin (3 θ) \sin (5 θ)$ as a sum containing only sines or only cosines.

Rewrite over a common denominator:

$\frac{\sin (θ) + \cos (θ)}{\cos (θ)} + \frac{\sin (θ) + \cos (θ)}{\sin (θ)}$

Establish each identity.

$\frac{\sin θ + \cos θ}{\sin θ} - \frac{\cos θ - \sin θ}{\cos θ} = s e c θ c s c θ$

See all solutions

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