What is the general solution for -sin(2θ)-cos(4θ)=0 ?

The general solution for -sin(2θ)-cos(4θ)=0 is θ=(pi+4pin)/4 ,θ=(7pi+12pin)/(12),θ=(11pi+12pin)/(12)

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-sin(2θ)-cos(4θ)=0

- sin (2 θ) - cos (4 θ) = 0

θ = \frac{π + 4 πn}{4}, θ = \frac{7 π + 12 πn}{12}, θ = \frac{11 π + 12 πn}{12}

Solution steps

- sin (2 θ) - cos (4 θ) = 0

Let:

u = 2 θ

- sin (u) - cos (2 u) = 0

Rewrite using trig identities

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

Solve by substitution

sin (u) = 1, sin (u) = - \frac{1}{2}

sin (u) = 1 : u = \frac{π}{2} + 2 πn

sin (u) = - \frac{1}{2} : u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

Combine all the solutions

u = \frac{π}{2} + 2 πn, u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

Substitute back

u = 2 θ

2 θ = \frac{π}{2} + 2 πn : θ = \frac{π + 4 πn}{4}

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π + 12 πn}{12}

2 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π + 12 πn}{12}

θ = \frac{π + 4 πn}{4}, θ = \frac{7 π + 12 πn}{12}, θ = \frac{11 π + 12 πn}{12}

5 tan (θ) + 9 = 0

3 sin (θ) = 3 sin (θ) + 3 cos (θ) + 3 sin (θ)

2 sin^{2} (x) - 3 sin (x) = 2

tan (x + π) + 2 sin (x + π) = 0, 0 \leq x \leq 2 π

sin (a) + 3 cos (a) = 0

What is the general solution for -sin(2θ)-cos(4θ)=0 ?
The general solution for -sin(2θ)-cos(4θ)=0 is θ=(pi+4pin)/4 ,θ=(7pi+12pin)/(12),θ=(11pi+12pin)/(12)