sin(2x)-sin(4x)+sin(6x)=0

Solution

sin (2 x) - sin (4 x) + sin (6 x) = 0

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{π + 6 πn}{6}, x = \frac{5 π + 6 πn}{6}, x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}

Solution steps

sin (2 x) - sin (4 x) + sin (6 x) = 0

Let:

u = 2 x

sin (u) - sin (2 u) + sin (3 u) = 0

Rewrite using trig identities

4 sin (u) - 4 sin^{3} (u) - 2 cos (u) sin (u) = 0

Factor

4 sin (u) - 4 sin^{3} (u) - 2 cos (u) sin (u) : 2 sin (u) (2 - 2 sin^{2} (u) - cos (u))

2 sin (u) (2 - 2 sin^{2} (u) - cos (u)) = 0

Solving each part separately

sin (u) = 0 or 2 - 2 sin^{2} (u) - cos (u) = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

2 - 2 sin^{2} (u) - cos (u) = 0 : u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn, u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Substitute back

u = 2 x

2 x = 2 πn : x = πn

2 x = π + 2 πn : x = \frac{π + 2 πn}{2}

2 x = \frac{π}{3} + 2 πn : x = \frac{π + 6 πn}{6}

2 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π + 6 πn}{6}

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

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

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{π + 6 πn}{6}, x = \frac{5 π + 6 πn}{6}, x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}

2 csc (x) - 3 = 0

cos (θ) = \frac{8}{12}

sin (x) sin (2 x) = 0

sin (x) = 1.2

9 cos^{2} (θ) = - 10 cos (θ) - 1