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

The general solution for 2cos^2(x)+sin(-x)-1=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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

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

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

Solution steps

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

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

cos (x) = \frac{- 1}{4}

tan (θ) = \frac{7}{2}

2 \cdot sin (2 x + 1 5^{\circ}) = 1

cos (2 x) + 3 cos (x) + 2 = 0, \frac{π}{2} \leq x \leq π

2 cos (a) = 3

What is the general solution for 2cos^2(x)+sin(-x)-1=0 ?
The general solution for 2cos^2(x)+sin(-x)-1=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin