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

The general solution for 2sin^2(x)+3sin(x)+1=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6 ,x=(3pi)/2

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

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

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{3 π}{2}

Solution steps

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

Solve by substitution

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

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{2}

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{3 π}{2}

cos (3 x) - 1 = sin (3 x)

sin (\frac{x}{3}) = \frac{2}{2}

2 cos (3 x) = - 3

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

- tan (x) = 1

What is the general solution for 2sin^2(x)+3sin(x)+1=0,0<= x<= 2pi ?
The general solution for 2sin^2(x)+3sin(x)+1=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6 ,x=(3pi)/2