How do I solve: cos2(x+pi/6)=1/√2 if the 2 would be before cos I think it would be easy but.. when it after I dont know what do to!

if I would have to guess I think I would do something like this: cos(2x+2pi/12)=1/√2 <=> 2x+2pi/12=pi/4 <=> 2x = (3pi/12)-(2pi/12) <=> 2x = pi/12 <=> x = pi/24
to me it looks about right since it should be a positive value.. so my final answer = +- pi/24 + n*2pi

pls correct me if im wrong

This post was edited by bomben on Oct 16 2013 02:05pm

which of these is your problem?

the bottom one u got there

limiting to angles in range [0,2pi]


This post was edited by Azrad on Oct 17 2013 12:23pm

Traditionally you should restrict your solutions so that you get values of x in the range [0, 2pi], rather than making sure 2x + pi/3 is in the desired range.

Yeah, I guess I stopped early, anyway those are the 2 unique reference angles

Convert them both to positive angles, and find their partners (by adding or subtracting pi) in the range[0,2pi]

x = 17pi/24, 23pi/24, 41pi/24, 47pi/24

This post was edited by Azrad on Oct 18 2013 12:05am