Chapter 7: Q. 91 (page 487)

solve each equation on the interval $0 \leq θ < 2 π$
$\sin θ - \sqrt{3} \cos θ = 1$

Short Answer

Expert verified

$θ = (n π + \frac{π}{3}) + {(- 1)}^{(n)} \frac{π}{6}$

Step by step solution

Step 1.Given information is

a trignometric equation $\sin θ - \sqrt{3} \cos θ = 1$

Step 2.multiplying both the side by 12, we get12sinθ-32cosθ=12cosπ3sinθ-sinπ3cosθ=sinπ6sinθ-π3=sinπ6

we know that

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

Step 3.the general solution we get is

$θ = n π + {(- 1)}^{n} \frac{π}{6} + \frac{π}{3}$

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solve each equation on the interval $0 \leq θ < 2 π$
$\sin θ - \sqrt{3} \cos θ = 1$

Short Answer

Step by step solution

Step 1.Given information is

Step 2.multiplying both the side by 12, we get12sinθ-32cosθ=12cosπ3sinθ-sinπ3cosθ=sinπ6sinθ-π3=sinπ6

Step 3.the general solution we get is

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solve each equation on the interval 0≤θ<2πsinθ-3cosθ=1

Short Answer

Step by step solution

Step 1.Given information is

Step 2.multiplying both the side by 12, we get12sinθ-32cosθ=12cosπ3sinθ-sinπ3cosθ=sinπ6sinθ-π3=sinπ6

Step 3.the general solution we get is

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

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Study anywhere. Anytime. Across all devices.

solve each equation on the interval $0 \leq θ < 2 π$
$\sin θ - \sqrt{3} \cos θ = 1$