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

The general solution for sin^2(x)+2sin(x)-1=0 is x=0.42707…+2pin,x=pi-0.42707…+2pin

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

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

x = 0.42707 \dots + 2 πn, x = π - 0.42707 \dots + 2 πn

Solution steps

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

Solve by substitution

sin (x) = - 1 + 2, sin (x) = - 1 - 2

sin (x) = - 1 + 2 : x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

sin (x) = - 1 - 2 :

No Solution

Combine all the solutions

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

Show solutions in decimal form

x = 0.42707 \dots + 2 πn, x = π - 0.42707 \dots + 2 πn

1 = cos^{3} (x)

sin (θ) = \frac{150 N}{212.6 N} (0.9063)

2 sin^{2} (x) = sin (x), 0 \leq x \leq 2 π

3 sin (x) + 4 = 5 cos (x), [0.2 π]

cos^{2} (x) + cos^{2} (2 x) = 1

What is the general solution for sin^2(x)+2sin(x)-1=0 ?
The general solution for sin^2(x)+2sin(x)-1=0 is x=0.42707…+2pin,x=pi-0.42707…+2pin