Chapter 7: Q. 8 (page 464)

All the solutions of the equation $\sin θ = \frac{1}{2}$ are ____.

Short Answer

Expert verified

All the solutions of the equation $\sin θ = \frac{1}{2}$ are $\{θ | θ = \frac{π}{6} + 2 πk, θ = \frac{5 π}{6} + 2 πk, k is any integer\}$ .

Step by step solution

Step 1. Given Information

All the solutions of the equation $\sin θ = \frac{1}{2}$ are ____.

Step 2. The period of the sine function is 2π.

In the interval $[0, 2 π)$ , there are two angles $θ$ for which

So all the solution of $\sin θ = \frac{1}{2}$ are width="341">θ|θ=π6+2πk,θ=5π6+2πk,kisanyinteger

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All the solutions of the equation $\sin θ = \frac{1}{2}$ are ____.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. The period of the sine function is 2π.

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All the solutions of the equation sinθ=12are ____.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. The period of the sine function is 2π.

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Most popular questions from this chapter

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All the solutions of the equation $\sin θ = \frac{1}{2}$ are ____.