What is the general solution for 2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi ?

The general solution for 2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6

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2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi

2 sin^{2} (x) - 5 sin (x) + 2 = 0, 0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

2 sin^{2} (x) - 5 sin (x) + 2 = 0, 0 \leq x \leq 2 π

Solve by substitution

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

sin (x) = 2, 0 \leq x \leq 2 π :

No Solution

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}

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

sec (θ) = sin^{2} (θ) sec (θ)

sin (2 θ) + sin (θ) = 0, 0 \leq θ \leq 2 π

1 + cos (x) = 2 - cos (x)

cos^{2} (x) - 3 cos (x) - 4 = 0

What is the general solution for 2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi ?
The general solution for 2sin^2(x)-5sin(x)+2=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6