What is the general solution for sin(2x)=-sqrt(2)sin(x) ?

The general solution for sin(2x)=-sqrt(2)sin(x) is x=2pin,x=pi+2pin,x=(3pi)/4+2pin,x=(5pi)/4+2pin

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sin(2x)=-sqrt(2)sin(x)

sin (2 x) = - 2 sin (x)

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Solution steps

sin (2 x) = - 2 sin (x)

Subtract

- 2 sin (x)

from both sides

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

Rewrite using trig identities

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

Factor

sin (x) 2 + 2 cos (x) sin (x) : sin (x) (2 + 2 cos (x))

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

2 + 2 cos (x) = 0 : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

10 sec^{2} (x) + 5 tan^{2} (x) - 15 = 0

sin (2 x) + 1 = 0

cos (2 x) cos (x) + sin (2 x) sin (x) = 1

cos (2 θ) + 10 sin^{2} (θ) = 7

4 cos (2 t) + 1 = 3

What is the general solution for sin(2x)=-sqrt(2)sin(x) ?
The general solution for sin(2x)=-sqrt(2)sin(x) is x=2pin,x=pi+2pin,x=(3pi)/4+2pin,x=(5pi)/4+2pin