46) Determine solutions for each equation in the interval 0 ≤ x ≤ 2π, to the nearest hundredth of a radian.
Give exact answers where possible.
c) 2 sin x = −√3
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.