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

The general solution for 2cos^2(x)-sin(x)=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin

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

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

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = - \frac{1 + 17}{4}, sin (x) = \frac{17 - 1}{4}

sin (x) = - \frac{1 + 17}{4} :

No Solution

sin (x) = \frac{17 - 1}{4} : x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Combine all the solutions

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Show solutions in decimal form

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

4 + 4 sin (x) = 8 cos (x)

- 3 tan^{2} (θ) - 4 tan (θ) = - 4

(tan (x) + 1) (3 cot (x) + 1) = 0

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

θ, x = r cos (θ)

What is the general solution for 2cos^2(x)-sin(x)=0 ?
The general solution for 2cos^2(x)-sin(x)=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin