What is the general solution for 14sin^2(x)+21sin(x)+7=0 ?

The general solution for 14sin^2(x)+21sin(x)+7=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin

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14sin^2(x)+21sin(x)+7=0

14 sin^{2} (x) + 21 sin (x) + 7 = 0

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

Solution steps

14 sin^{2} (x) + 21 sin (x) + 7 = 0

Solve by substitution

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

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

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

2 sec (θ) + 6 = 10

2 sin^{2} (θ) = 3 sin (θ) - 1

cot (θ) = - 5

2 + 2 sin (x) = 0

2 sin (2 x) = 1, 0 < x < 2 π

What is the general solution for 14sin^2(x)+21sin(x)+7=0 ?
The general solution for 14sin^2(x)+21sin(x)+7=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin