SOLUTION: Please help me solve these 3 equations for solutions from [0, 2pi]: 1.) {{{2cos^2(2x)-cos2x=0}}} 2.) {{{2sin^2(x)-sin(x)-1=0}}} 3.) {{{tan^2(3x)=-tan(3x)}}} Thank you!!

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me solve these 3 equations for solutions from [0, 2pi]: 1.) {{{2cos^2(2x)-cos2x=0}}} 2.) {{{2sin^2(x)-sin(x)-1=0}}} 3.) {{{tan^2(3x)=-tan(3x)}}} Thank you!!      Log On


   

Question 983506: Please help me solve these 3 equations for solutions from [0, 2pi]:
1.) 2cos%5E2%282x%29-cos2x=0
2.) 2sin%5E2%28x%29-sin%28x%29-1=0
3.) tan%5E2%283x%29=-tan%283x%29
Thank you!!!
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!

1. 2cos%5E2%282x%29 - cos%282x%29 = 0.

Introduce new variable   y = cos%282x%29.

Then your equation will be

2y%5E2 - y = 0.

One solution is  y = 0,  i.e.   cos%282x%29 = 0,  which gives   2x = pi%2F2  or   2x = 3pi%2F2.
Please complete this case yourself.

Next,  if  y is not zero,  then you can reduce the original equation to the form

2y - 1 = 0,   i.e.   2cos%282x%29 = 1.

It gives you   cos%282x%29 = 1%2F2,  and then   2x = pi%2F3  or   2x = 2pi+-+pi%2F3 = 5pi%2F3.

You can complete this case yourself.


The remaining equations can be solved using the same trick with introduction of a new variable.

You can do it yourself.

Good luck !!!