46) Determine solutions for each equation in the interval 0 ≤ x ≤ 2π, to the nearest hundredth of a radian.
Give exact answers where possible.
d) 2 sin x cos x − cos x = 0
To solve the equation 2 sin x cos x − cos x = 0, we can factor out cos x:
cos x(2 sin x - 1) = 0
Setting each factor equal to zero gives us two separate equations:
cos x = 0
2 sin x - 1 = 0
For the first equation, cos x = 0, we know that cos x is equal to zero at x = π/2 and x = 3π/2 in the interval 0 ≤ x ≤ 2π.
For the second equation, 2 sin x - 1 = 0, we solve for sin x:
2 sin x = 1
sin x = 1/2
The value sin x = 1/2 is true for x = π/6 and x = 5π/6 in the interval 0 ≤ x ≤ 2π.
Therefore, the solutions for the equation 2 sin x cos x − cos x = 0 in the interval 0 ≤ x ≤ 2π are:
x = π/2, 3π/2, π/6, and 5π/6.