What is the general solution for sin(2x)+2cos^2(x)=0 ?

The general solution for sin(2x)+2cos^2(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin,x=(3pi)/4+pin

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sin(2x)+2cos^2(x)=0

sin (2 x) + 2 cos^{2} (x) = 0

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{3 π}{4} + πn

Solution steps

sin (2 x) + 2 cos^{2} (x) = 0

Rewrite using trig identities

2 cos^{2} (x) + 2 cos (x) sin (x) = 0

Factor

2 cos^{2} (x) + 2 cos (x) sin (x) : 2 cos (x) (cos (x) + sin (x))

2 cos (x) (cos (x) + sin (x)) = 0

Solving each part separately

cos (x) = 0 or cos (x) + sin (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) + sin (x) = 0 : x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{3 π}{4} + πn

sin (x) = - (\frac{1}{2})

x, y - cos (x) = sin (y)

cos (x) = 0, 0 \leq x < 2 π

cos^{3} (x) - 2 cos^{2} (x) + cos (x) - 2 = 0

cos^{2} (x) - sin^{2} (x) = 1 + sin (x)

What is the general solution for sin(2x)+2cos^2(x)=0 ?
The general solution for sin(2x)+2cos^2(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin,x=(3pi)/4+pin