Question

Find the angle θ (in radians) such that 5cos⁡(2θ)−1=2, so that

7π≤θ≤8π, and so that θ is in the third quadrant.

Answer 1

Explanation:

5×cos(2x) - 1 = 2

5×cos(2x) = 3

cos(2x) = 3/5

cos(2x) = 2×cos²(x) - 1

2×cos²(x) - 1 = 3/5

2×cos²(x) = 3/5 + 1 = 3/5 + 5/5 = 8/5

cos²(x) = 8/5 / 2 = 8/10 = 4/5

cos(x) = 2/sqrt(5)

when x (theta) is in the third quadrant, then cos(x) is negative. that means we need the negative square root of 5.

cos(x) = -2/sqrt(5) = -0.894427191...

x = 2.677945045 + 2pi×n

or

x = 2pi - 2.677945045... + 2pi×n

= 3.605240263... + 2pi×n

again, for the third quadrant the only possible solution (x > pi) is

x = 3.605240263... + 2pi×n

for x to be between 7pi and 8pi, n must be 3

x (theta) = 3.605240263... + 2pi×3 =

= 3.605240263... + 6pi =

= 22.45479618...