SOLUTION: Find all x in the interval [0,2pi] a) sin^2(2x)+3sin^2(x)=3 b) (cos(x))/(2sin(pi/2))-sin(x))≥0

Algebra ->  Trigonometry-basics -> SOLUTION: Find all x in the interval [0,2pi] a) sin^2(2x)+3sin^2(x)=3 b) (cos(x))/(2sin(pi/2))-sin(x))≥0      Log On


   

Question 990313: Find all x in the interval [0,2pi]
a) sin^2(2x)+3sin^2(x)=3
b) (cos(x))/(2sin(pi/2))-sin(x))≥0
Answer by ikleyn(52797) About Me  (Show Source):
You can put this solution on YOUR website!
.
I will help you to solve the equation a)

sin%5E2%282x%29 + 3sin%5E2%28x%29 = 3.                             (1)

From  Trigonometry,  there is an identity

sin%282x%29 = 2sin%28x%29%2Acos%28x%29.

Substitute it into your equation.  You will get

4sin%5E2%28x%29%2Acos%5E2%28x%29 + 3sin%5E2%28x%29 = 3.                 (2)

Recall that cos%5E2%28x%29 = 1-sin%5E2%28x%29.

Substitute it into your equation (2).  You will get

4sin%5E2%28x%29%2A%281-sin%5E2%28x%29%29 + 3sin%5E2%28x%29 = 3.          (3)

To solve this equation, introduce new variable  y = sin%5E2%28x%29.  Then you will get the equation

4y%2A%281-y%29 + 3y = 3                                        (4)

instead of (3).  Simplify  (4):

4y%5E2+-+7y+%2B3 = 0.

Solve this quadratic equation.  Use the quadratic formula.  Its roots are  y = 1  and  y = 3%2F4.

So,  now you need to solve two equations:

1) sin%5E2%28x%29 = 1,  or  sin%28x%29 = +/- 1.

and

2) sin%5E2%28x%29 = 3%2F4,  or  sin%28x%29 = +/- sqrt%283%29%2F2.

Please do it yourself.