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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. 95 (page 487)

solve each equation on the interval
$0 鈮� 胃 < 2 蟺$
$\tan (胃) + \sqrt{3} = \frac{1}{\cos (胃)}$

Short Answer

Expert verified

$胃 = n 蟺 + {(- 1)}^{n} \frac{蟺}{3} - \frac{蟺}{3}$

Step by step solution

01

Step 1.The information given istanθ+3=1cosθ

a trigonometric equation.

02

Step 2. Solving it we get,

$\tan (胃) + \sqrt{3} = \frac{1}{\cos (胃)} \frac{\sin 胃}{\cos 胃} + \sqrt{3} = \frac{1}{\cos 胃} \frac{\sin 胃 + \sqrt{3} \cos 胃}{\cos 胃} = \frac{1}{\cos 胃} \sin 胃 + \sqrt{3} \cos 胃 = 1$

03

Step 3. Multiplying both the side by 12

$\frac{1}{2} \sin 胃 + \frac{\sqrt{3}}{2} \cos 胃 = \frac{1}{2} \cos (\frac{蟺}{3}) \sin (胃) + \sin (\frac{蟺}{3}) \cos (胃) = \sin (\frac{蟺}{3}) \sin (胃 + \frac{蟺}{3}) = \sin (\frac{蟺}{3})$

04

Step 4. The general solution we get is

$胃 = n 蟺 + {(- 1)}^{n} \frac{蟺}{3} - \frac{蟺}{3}$

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Most popular questions from this chapter

If $z = \tan \frac{伪}{2}$ , show that localid="1646416257885" $\cos 伪 = \frac{1 - z^{2}}{1 + z^{2}}$ .

True or False : In establishing an identity, it is often easiest to just multiply both sides by a well-chosen nonzero expression involving the variable.

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

Establish each identity.

$\frac{1 + \cos 胃 + \sin 胃}{1 + \cos 胃 - \sin 胃} = s e c 胃 + \tan 胃$

Establish each identity.

${(a \sin 胃 + b \cos 胃)}^{2} + {(a \cos 胃 - b \sin 胃)}^{2} = a^{2} + b^{2}$

See all solutions

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