What is the general solution for 4sin^2(x)-5sin(x)+1=0 ?

The general solution for 4sin^2(x)-5sin(x)+1=0 is x= pi/2+2pin,x=0.25268…+2pin,x=pi-0.25268…+2pin

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4sin^2(x)-5sin(x)+1=0

4 sin^{2} (x) - 5 sin (x) + 1 = 0

x = \frac{π}{2} + 2 πn, x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn

Solution steps

4 sin^{2} (x) - 5 sin (x) + 1 = 0

Solve by substitution

sin (x) = 1, sin (x) = \frac{1}{4}

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = \frac{1}{4} : x = arcsin (\frac{1}{4}) + 2 πn, x = π - arcsin (\frac{1}{4}) + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = arcsin (\frac{1}{4}) + 2 πn, x = π - arcsin (\frac{1}{4}) + 2 πn

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn

tan (x) = - 0.6293

2 cos (\frac{2 x}{3}) - 3 = 0

- sin (2 x) - sin (x) = 0

sin^{2} (t) + 2 cos (t) = 1

6 tan (θ) - 2 cot (θ) = 0

What is the general solution for 4sin^2(x)-5sin(x)+1=0 ?
The general solution for 4sin^2(x)-5sin(x)+1=0 is x= pi/2+2pin,x=0.25268…+2pin,x=pi-0.25268…+2pin