Chapter 7: Q. 95 (page 487)

solve each equation on the interval
$0 \leq θ < 2 π$
$\tan (θ) + \sqrt{3} = \frac{1}{\cos (θ)}$

Short Answer

Expert verified

$θ = n π + {(- 1)}^{n} \frac{π}{3} - \frac{π}{3}$

Step by step solution

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

a trigonometric equation.

Step 2. Solving it we get,

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

Step 3. Multiplying both the side by 12

$\frac{1}{2} \sin θ + \frac{\sqrt{3}}{2} \cos θ = \frac{1}{2} \cos (\frac{π}{3}) \sin (θ) + \sin (\frac{π}{3}) \cos (θ) = \sin (\frac{π}{3}) \sin (θ + \frac{π}{3}) = \sin (\frac{π}{3})$

Step 4. The general solution we get is

$θ = n π + {(- 1)}^{n} \frac{π}{3} - \frac{π}{3}$

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solve each equation on the interval
$0 \leq θ < 2 π$
$\tan (θ) + \sqrt{3} = \frac{1}{\cos (θ)}$

Short Answer

Step by step solution

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

Step 2. Solving it we get,

Step 3. Multiplying both the side by 12

Step 4. The general solution we get is

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solve each equation on the interval0≤θ<2πtanθ+3=1cosθ

Short Answer

Step by step solution

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

Step 2. Solving it we get,

Step 3. Multiplying both the side by 12

Step 4. The general solution we get is

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Applied Mathematics

Statistics

Theoretical and Mathematical Physics

Probability and Statistics

Pure Maths

Study anywhere. Anytime. Across all devices.

solve each equation on the interval
$0 \leq θ < 2 π$
$\tan (θ) + \sqrt{3} = \frac{1}{\cos (θ)}$