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

The general solution for sin(x)-2cos^2(x)=-1 is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

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

Subtract

- 1

from both sides

sin (x) - 2 cos^{2} (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{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

sin (A) = \frac{6.347}{12.7}

sec (x) = 3 cos (x) - tan (x)

arctan (x^{2} - 2 x) = 0

sin (x^{\circ}) = \frac{6.57 \cdot 1 0 ^{- 7}}{1.8 \cdot 1 0 ^{- 6}}

What is the general solution for sin(x)-2cos^2(x)=-1 ?
The general solution for sin(x)-2cos^2(x)=-1 is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin