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Question
math 195 section 9.5 exercises
- solve each trigonometric equation, for all values of $\theta$ on the interval $0 \leq \theta \leq 2\pi$.
(a) $2\sin\theta=-\sqrt{2}$
$\frac{2\sin\theta}{2}=\frac{-\sqrt{2}}{2}$
$\sin\theta=-\frac{\sqrt{2}}{2}$
$\sin\theta < 0$ (qiii, qiv)
$\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}-$ qiii
$\theta=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}\to$ qiv
$\theta=\frac{5\pi}{4},\frac{7\pi}{4}$
(b) $2\sin\theta=\sqrt{3}$
$\frac{2\sin\theta}{2}=\frac{\sqrt{3}}{2}$
$\sin\theta=\frac{\sqrt{3}}{2}$
$\frac{\pi}{3}$
(c) $2\cos\theta=1$
$\frac{2\cos\theta}{2}=\frac{1}{2}$
$\cos\theta=\frac{1}{2}$
(d) $2\cos\theta=-\sqrt{2}$
$\frac{2\cos\theta}{2}=\frac{-\sqrt{2}}{2}$
$\cos\theta=-\frac{\sqrt{2}}{2}$
Step1: Solve \(2\sin\theta =-\sqrt{2}\)
Divide both sides by \(2\): \(\sin\theta=-\frac{\sqrt{2}}{2}\).
The reference angle \(\theta_{ref}=\frac{\pi}{4}\).
Since \(\sin\theta<0\), \(\theta\) is in QIII and QIV.
For QIII: \(\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\).
For QIV: \(\theta = 2\pi-\frac{\pi}{4}=\frac{7\pi}{4}\).
Step2: Solve \(2\sin\theta=\sqrt{3}\)
Divide both sides by \(2\): \(\sin\theta=\frac{\sqrt{3}}{2}\).
The reference angle \(\theta_{ref}=\frac{\pi}{3}\).
Since \(\sin\theta>0\), \(\theta\) is in QI and QII.
For QI: \(\theta=\frac{\pi}{3}\).
For QII: \(\theta=\pi-\frac{\pi}{3}=\frac{2\pi}{3}\).
Step3: Solve \(2\cos\theta = 1\)
Divide both sides by \(2\): \(\cos\theta=\frac{1}{2}\).
The reference angle \(\theta_{ref}=\frac{\pi}{3}\).
Since \(\cos\theta>0\), \(\theta\) is in QI and QIV.
For QI: \(\theta=\frac{\pi}{3}\).
For QIV: \(\theta = 2\pi-\frac{\pi}{3}=\frac{5\pi}{3}\).
Step4: Solve \(2\cos\theta=-\sqrt{2}\)
Divide both sides by \(2\): \(\cos\theta=-\frac{\sqrt{2}}{2}\).
The reference angle \(\theta_{ref}=\frac{\pi}{4}\).
Since \(\cos\theta<0\), \(\theta\) is in QII and QIII.
For QII: \(\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}\).
For QIII: \(\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\).
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(a) \(\theta=\frac{5\pi}{4},\frac{7\pi}{4}\)
(b) \(\theta=\frac{\pi}{3},\frac{2\pi}{3}\)
(c) \(\theta=\frac{\pi}{3},\frac{5\pi}{3}\)
(d) \(\theta=\frac{3\pi}{4},\frac{5\pi}{4}\)