What is the general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<= 180 ?

The general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<= 180 is x=90,x=30,x=150

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2sin^2(x)-3sin(x)+1=0,0<= x<= 180

2 sin^{2} (x) - 3 sin (x) + 1 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

x = 9 0^{\circ}, x = 3 0^{\circ}, x = 15 0^{\circ}

Solution steps

2 sin^{2} (x) - 3 sin (x) + 1 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

Solve by substitution

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

sin (x) = 1, 0 \leq x \leq 18 0^{\circ} : x = 9 0^{\circ}

sin (x) = \frac{1}{2}, 0 \leq x \leq 18 0^{\circ} : x = 3 0^{\circ}, x = 15 0^{\circ}

Combine all the solutions

x = 9 0^{\circ}, x = 3 0^{\circ}, x = 15 0^{\circ}

sin^{2} (θ) + cos (θ) = 2

cot (2 x) - 1 = 0

tan (2 θ) - tan (θ) = 0, 0 < θ < 2 π

- 8 cos (x) = - 6 sec (x) + 8

cos (2 θ) + sin (θ) = 1

What is the general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<= 180 ?
The general solution for 2sin^2(x)-3sin(x)+1=0,0<= x<= 180 is x=90,x=30,x=150