SOLUTION: please help me solve this equation. Solve the equation for x over the interval [0,2pi) {{4cos2x=squareroot3}}

Question 682764: please help me solve this equation. Solve the equation for x over the interval [0,2pi) {{4cos2x=squareroot3}}
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
please help me solve this equation. Solve the equation for x over the interval [0,2pi) {{4cos2x=squareroot3}}
4cos(2x)=√3
cos(2x)=√3/4
Identity: cos(2x)=1-2sin^2x
1-2sin^2x=√3/4
2sin^2x=1-√3/4=(4-√3)/4
sin^2x=(4-√3)/8≈.2835
sinx=√.2835=≈.5324
using calculator set to radians
x=0.5685 and 2.5801 (in Q1 and Q2 where sin>0)
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