46) Determine solutions for each equation in the interval 0 ≤ x ≤ 2π, to the nearest hundredth of a radian.

Question

Give exact answers where possible.

c) 2 sin x = −√3

Answer 1

To solve the equation 2 sin x = −√3, we can start by dividing both sides by 2:

sin x = -√3/2

We know that the sine function is negative in the third and fourth quadrants. In the third quadrant, the reference angle that has a sine value of -1/2 is π/6. So one solution in the interval 0 ≤ x ≤ 2π is:

x = π + π/6 = 7π/6 radians

In the fourth quadrant, the reference angle that has a sine value of -1/2 is also π/6. So another solution in the interval 0 ≤ x ≤ 2π is:

x = 2π - π/6 = 11π/6 radians

Therefore, the solutions to the equation 2 sin x = −√3 in the interval 0 ≤ x ≤ 2π are approximately 7π/6 and 11π/6 radians to the nearest hundredth of a radian.