Trigonometric Equations?

Question

What is the easiest method for solving trigonometric equations?

I have two questions below. I am looking for the easiest method to apply. As you answer the two questions, please show me step by step what you are doing.

(1) Give the general solution to 2cos2x + 1 = 0

NOTE: For question (1), by "general solution" the author means to include the term 2kpi with each value found for x, where k = any integer.

(2) Give all solutions to the equation 2cos^2(3x) - 1 = 0 on the interval [0, 2pi).

Thanks

Answer 1

(1) cos 2x = - 1/2
2x = (2k+1)*pi +- pi/3
x = (k + 0.5)*pi +- pi/6
(2) 2cos^2(3x) - 1 = 0
This can be written as:
0 = 1 - 2cos^2(3x)
= [cos^2(3x) + sin^2(3x)] - 2cos^2(3x)
= - [cos^2(3x) - sin^2(3x)]
= - cos 2(3x) = - cos(6x)
Thus: 6x = 2k(pi) +- pi/2
x = k pi/3 +- pi/12 = (2k + 1)pi/12
Explicitely, x = pi/12, pi/4, 5 pi/12, 7pi/12, 3pi/4, and 11pi/12