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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 72 (page 504)

Solve the equation on the interval $0 鈮� 胃鈮� 2 蟺$ .
$2 \sin^{2} 胃 - 3 \sin 胃 + 1 = 0$

Short Answer

Expert verified

On solving the equation, we get,

$胃鈭� \{\frac{蟺}{6}, \frac{蟺}{2}, \frac{5 蟺}{6}\}$

Step by step solution

01

Step 1. Given information.

Consider the given question,

$2 \sin^{2} 胃 - 3 \sin 胃 + 1 = 0 0 鈮� 胃鈮� 2 蟺$

On rewriting the equation,

$2 \sin^{2} 胃 - 2 \sin 胃 - \sin 胃 + 1 = 0 2 \sin 胃 (\sin 胃 - 1) - (\sin 胃 - 1) = 0 (\sin 胃 - 1) (2 \sin 胃 - 1) = 0$

Then,

$\sin 胃 - 1 = 0 \sin 胃 = 1$

Also,

$2 \sin 胃 - 1 = 0 \sin 胃 = \frac{1}{2}$

02

Step 2. Solve the equation for the given interval.

Consider the given question,

$\sin 胃 = 1 \sin 胃 = \frac{1}{2} 0 鈮� 胃鈮� 2 蟺$

Solving $\sin 胃 = 1$ for the given interval,

role="math" localid="1646747872475" $\sin 胃 = 1 胃 = \sin^{- 1} (1) 胃 = \frac{蟺}{2}$

Solving $\sin 胃 = \frac{1}{2}$ for the given interval,

$胃 = \sin^{- 1} (\frac{1}{2}) 胃 = \frac{蟺}{6}, \frac{5 蟺}{6}$

Therefore, the solution set is $胃鈭� \{\frac{蟺}{6}, \frac{蟺}{2}, \frac{5 蟺}{6}\}$

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Most popular questions from this chapter

If $\tan 伪 = x + 1$ and role="math" localid="1646425111199" $\tan 尾 = x - 1,$ show that

role="math" localid="1646426027733" $2 c o t (伪 - b) = x^{2}$

True or False:

$\tan 2 胃 + \tan 2 胃 = \tan (4 胃)$

Establish each identity.

$\frac{{(2 \cos^{2} 胃 - 1)}^{2}}{\cos^{4} 胃 - \sin^{4} 胃} = 1 - 2 \sin^{2} 胃$

Rewrite in terms of sine and cosine functions: $c o t (胃) s e c (胃)$

Multiply: $\frac{\cos (胃)}{1 - \sin (胃)} \frac{1 + \sin (胃)}{1 + \sin (胃)}$

See all solutions

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