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

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

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

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

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

Solution steps

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

cosh (4 x) = 16 sinh (x) + 1

cos (θ) = (\frac{3}{2})

cos (x + \frac{3 π}{4}) + cos (x - \frac{3 π}{4}) = 1

5 - sin (θ) = cos (2 θ)

8 sin (θ) + 43 = 0

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