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

The general solution for 2sin^2(x)+2cos(x)-1=0 is x=1.94553…+2pin,x=-1.94553…+2pin

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

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

x = 1.94553 \dots + 2 πn, x = - 1.94553 \dots + 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - \frac{- 1 + 3}{2} : x = arccos (- \frac{- 1 + 3}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 3}{2}) + 2 πn

cos (x) = \frac{1 + 3}{2} :

No Solution

Combine all the solutions

x = arccos (- \frac{- 1 + 3}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 3}{2}) + 2 πn

Show solutions in decimal form

x = 1.94553 \dots + 2 πn, x = - 1.94553 \dots + 2 πn

cos (x + π) = sin (x)

2 cos^{2} (θ) tan (θ) - tan (θ) = 0

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

7 cos^{2} (θ) + 11 cos (θ) = cos (θ) - 3

cos^{2} (x) - 9 cos (x) - 1 = 0

What is the general solution for 2sin^2(x)+2cos(x)-1=0 ?
The general solution for 2sin^2(x)+2cos(x)-1=0 is x=1.94553…+2pin,x=-1.94553…+2pin