Question

How do you find all solutions in the interval [0,2pi) of the equation (cosx)2−2.6cosx−0.87=0?

Answer 1

Answer:

For (0, 2pi)

pi/3, pi, (5pi)/3

Explanation:

Use trig identity:

sin x = 2sin (x/2).cos (x/2)

Solve the equation:

cos (x/2) - 2sin (x/2).cos (x/2) = 0

cos (x/2)((1 - 2sin (x/2) = 0

a. cos x/2 = 0 --> x/2 = pi/2, and x/2 = (3pi)/2 -->

x/2 = pi/2 --> x = pi

x/2 = (3pi)/2 --> x = 3pi (rejected because out of area)

b. (1 - 2sin (x/2) = 0

Trig table and unit circle -->

sin (x/2) = 1/2 --> x/2 = pi/6 and x/2 = (5pi)/6

x/2 = pi/6 --> x = pi/3

x/2 = (5pi)/6 --> x = (5pi)/3

Answers for (0, 2pi) pi/3, pi, (5pi)/3#