What is the general solution for 2sin^2(x)+sin^3(x)-1=0 ?

The general solution for 2sin^2(x)+sin^3(x)-1=0 is x=(3pi)/2+2pin,x=0.66623…+2pin,x=pi-0.66623…+2pin

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2sin^2(x)+sin^3(x)-1=0

2 sin^{2} (x) + sin^{3} (x) - 1 = 0

x = \frac{3 π}{2} + 2 πn, x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

Solution steps

2 sin^{2} (x) + sin^{3} (x) - 1 = 0

Solve by substitution

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

sin (x) = - 1 : x = \frac{3 π}{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

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

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

x = \frac{3 π}{2} + 2 πn, x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

2 cos^{2} (x) = 3 cos (x) - 1

sin^{2} (x) - 4 sin (x) + 4 = 0

3 cos^{2} (x) - 10 cos (x) + 3 = 0

(m + 1) sin (x) + 2 - m = 0

sin^{5} (x) + sin (x) + 2 sin^{2} (x) = 1

What is the general solution for 2sin^2(x)+sin^3(x)-1=0 ?
The general solution for 2sin^2(x)+sin^3(x)-1=0 is x=(3pi)/2+2pin,x=0.66623…+2pin,x=pi-0.66623…+2pin