What is the general solution for sin^4(x)+sin^2(x)=sin^6(x) ?

The general solution for sin^4(x)+sin^2(x)=sin^6(x) is x=2pin,x=pi+2pin

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sin^4(x)+sin^2(x)=sin^6(x)

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

x = 2 πn, x = π + 2 πn

Solution steps

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

Solve by substitution

sin (x) = 0, sin (x) = \frac{1 + 5}{2}, sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{1 - 5}{2}, sin (x) = - \frac{1 - 5}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = \frac{1 + 5}{2} :

No Solution

sin (x) = - \frac{1 + 5}{2} :

No Solution

sin (x) = \frac{1 - 5}{2} : x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn

sin (x) = - \frac{1 - 5}{2} : x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn, x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

Since the equation is undefined for:

arcsin \frac{1 - 5}{2} + 2 πn, π + arcsin - \frac{1 - 5}{2} + 2 πn, arcsin - \frac{1 - 5}{2} + 2 πn, π + arcsin \frac{1 - 5}{2} + 2 πn

x = 2 πn, x = π + 2 πn

cos (a) = \frac{- 11}{14}

x, sin (\frac{x}{x}) = 0.7

5 cos^{2} (x) + sin^{2} (x) = 4

cos (u) - 1.5 sin^{2} (u) + 0.1667 = 0

3 sin (2 x - 1) = 1

What is the general solution for sin^4(x)+sin^2(x)=sin^6(x) ?
The general solution for sin^4(x)+sin^2(x)=sin^6(x) is x=2pin,x=pi+2pin