find all solutions of each equations

Question

a) cos2x=square root of 3/2

b)cos 2theta/3 =-1

c) cot 3 theta /2 = -square root of 3

d) sin (2x-pie/4) = square root of 2/2

Answer 1

You should have a common method of attack for these kind of problems
What do you find difficult about them?

I will do the first, you try the same approach for the others.

cos 2Ø = √3/2
since the cosine is positive, we know that 2Ø must be in either quadrant I or quad V, by the CAST rule.
using my calculator, (or knowing the ratio of the 3-060-90 triangle),
we know cos 30 = √3/2

so 2Ø = 30° in I or 2Ø = 360-30 = 330° in IV

then Ø = 15° or Ø = 165° as our initial solutions
which in radians would be π/12 or 11π/12

Since you wanted "all solutions" , we look at the period of cos 2Ø
since the period of cos 2Ø = 360/2 = 180° or π radians, we add/subtract integer multiples of 18° to each answer,
general solution:
Ø = 15° +180k° , or 165+ 180k , where k is an integer
in radians,
Ø = π/12 + kπ or 11π/12 + kπ