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

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

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

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

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

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

1 + sin (x) - 2 sin^{2} (x) = 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{π}{2} + 2 πn

Combine all the solutions

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

sec (x) = 0

3 csc (x) - 2 = 0

cot (x) = 1

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

cot (x) = - 1

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