SOLUTION: solve: {{{2sin(2x)+cos(2x)+1=0}}} for {{{0<=x<=360}}}

Question 884078: solve: for
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
solve: for
***

square both sides
cos^2(2x)+2cos(2x)+1=4(1-cos^2(2x)
cos^2(2x)+2cos(2x)+1=4-4cos^2(2x)
5cos^2(2x)+2cos(2x)-3=0
(5cos(2x)-3)(cos(2x)+1)=0
..
(5cos(2x)-3)=0
cos(2x)=3/5
2x≈53.13˚ and 306.87
x≈26.57˚ and 153.43˚
reject, (extraneous roots)
..
Solution:
cos(2x)+1=0
cos(2x)=-1
2x=180˚
x=90˚

RELATED QUESTIONS
Solve ( 2sin^2x - cos^2x - 2 = 0 ) for ( 0� ≤ θ < 360� ). (answered by lwsshak3)

Solve 1+cos(2x)=2sin^2(2x) for x, where 0 ≤ to x < 2π (answered by lwsshak3)

Solve 1+cos(2x) = 2sin^2(2x) for x, where 0<= x < 2pie (answered by lwsshak3)

solve for 0<=x<=360... (answered by Theo)

Solve the equation for x, on the interval 0 is less than or equal to x and 2pi is greater (answered by Alan3354)

Solve for x on the interval given: a) 2sin^2x-5cosx+1=0 [0, pie] b) sinx/2=0 [-pie,... (answered by Alan3354)

solve for x E[0,2pi]... (answered by stanbon)

Solve to the nearest degree if needed, for 0≤x≤360� sin^2x=cosx... (answered by lwsshak3)

Cos 2x + cos x + 1= 0.... (answered by Alan3354,richard1234)