What is the general solution for 8cos(x)+4sin(2x)=0 ?

The general solution for 8cos(x)+4sin(2x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin

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8cos(x)+4sin(2x)=0

8 cos (x) + 4 sin (2 x) = 0

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

Solution steps

8 cos (x) + 4 sin (2 x) = 0

Rewrite using trig identities

8 cos (x) + 8 cos (x) sin (x) = 0

Factor

8 cos (x) + 8 cos (x) sin (x) : 8 cos (x) (sin (x) + 1)

8 cos (x) (sin (x) + 1) = 0

Solving each part separately

cos (x) = 0 or sin (x) + 1 = 0

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

sin (x) + 1 = 0 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

sin (x) = - \frac{4}{5}, cos (x) < 0, cos (\frac{x}{2})

1 + sin (x) - 2 sin^{2} (x) = 0

y, z = e^{- x} sin (x + 2 y)

3 sin (θ) - 2 sin^{2} (θ) = 0

3 tan (\frac{x}{3}) + 3 = 0

What is the general solution for 8cos(x)+4sin(2x)=0 ?
The general solution for 8cos(x)+4sin(2x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin