SOLUTION: Solve the equation for solutions in the interval 0 ≤ x < 2𝜋. Round approximate solutions to the nearest ten-thousandth. 2 sin^2 x − 1 = 0

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the equation for solutions in the interval 0 ≤ x < 2𝜋. Round approximate solutions to the nearest ten-thousandth. 2 sin^2 x − 1 = 0      Log On


   

Question 1178159: Solve the equation for solutions in the interval 0 ≤ x < 2𝜋. Round approximate solutions to the nearest ten-thousandth.
2 sin^2 x − 1 = 0
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
2sin^2(x)=1
sin^2(x)=(1/2)
sin(x)=+/- sqrt(2)/2=+/-0.7071
this is 45,135,225,315 degrees