Question

Find the possible values of x for which sin−1(x2−1)=31​π, giving your answers correct to 3 decimal places. (b) Solve the equation sin(2θ+31​π)=21​ for 0⩽θ⩽π, giving θ in terms of π in your answers.

Answer 1

The possible values of x for sin⁻¹(x²−1)=⅓π are approximately 1.366 and -1.366. For solving sin(2θ+⅓π)=½ within the range 0≩θ≩π, θ is approximately 0.262π.

Step-by-step explanation:

The first part of the question asks to find the possible values of x for which sin⁻¹(x²−1)=⅓π, which means the inverse sine of (x squared minus 1) is one third of pi. Since the reverse of the sine function, which is sine itself, must yield a value that's within the range of -1 to 1 (the domain of the inverse sine function), we solve the equation as follows:

After solving for x, we get x = ±√(1 + √3/2).

In the second part, we solve the equation sin(2θ+⅓π)=½ where 0≩θ≩π. The steps are:

The final equation for 2θ is 2θ = π/6 - ⅓π, solving for θ gives us θ = π/12 or approximately 0.262π.

So, the value of θ in terms of π is roughly 0.262π.