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

The general solution for 2cos^2(x)+5sin(x)+1=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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2cos^2(x)+5sin(x)+1=0

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

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

Solution steps

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

Rewrite using trig identities

3 - 2 sin^{2} (x) + 5 sin (x) = 0

Solve by substitution

sin (x) = - \frac{1}{2}, sin (x) = 3

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

sin (x) = 3 :

No Solution

Combine all the solutions

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

\frac{125}{sin ( 11 0 ^{\circ} )} = \frac{100}{sin ( b )}

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

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

cos (3 x) \cdot 3 = 0

16 sec^{2} (θ) - 25 = 0

What is the general solution for 2cos^2(x)+5sin(x)+1=0 ?
The general solution for 2cos^2(x)+5sin(x)+1=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin