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

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.Com

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) (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

RELATED QUESTIONS
Solve the equation for solutions in the interval 0 ≤ x < 2𝜋. Round approximate... (answered by Boreal,MathLover1)

Approximate to the nearest degree, the solutions of the equation in the interval [0�,... (answered by stanbon)

SOLVE EACH EQUATION, WHERE 0 DEGREES (< OR = TO) X < 360 DEGREES. ROUND APPROXIMATE... (answered by Edwin McCravy)

Approximate to the nearest degree, the solutions of the equation in the interval [0�,... (answered by lwsshak3)

Solve the equation 3x^2+2x-1=0 using the quadratic formula. Find the exact solutions.... (answered by MathLover1)

Solve x^2 - 5x - 1 = 0 using the Quadratic Formula. Find the exact solutions. Then... (answered by TimothyLamb)

round 399,218 to the nearest ten... (answered by ewatrrr)

round 376,415 to the nearest ten thousand (answered by Fombitz,akshay007)

Solve 3log2x=4. Round to the nearest... (answered by lwsshak3)