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

The general solution for 2cos(x)=sin^2(x) is x=1.14371…+2pin,x=2pi-1.14371…+2pin

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

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

x = 1.14371 \dots + 2 πn, x = 2 π - 1.14371 \dots + 2 πn

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = - 1 + 2, cos (x) = - 1 - 2

cos (x) = - 1 + 2 : x = arccos (- 1 + 2) + 2 πn, x = 2 π - arccos (- 1 + 2) + 2 πn

cos (x) = - 1 - 2 :

No Solution

Combine all the solutions

x = arccos (- 1 + 2) + 2 πn, x = 2 π - arccos (- 1 + 2) + 2 πn

Show solutions in decimal form

x = 1.14371 \dots + 2 πn, x = 2 π - 1.14371 \dots + 2 πn

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

2 sin^{2} (θ) - 5 sin (θ) - 3 = 0

4 cos^{2} (x) - 4 cos (x) - 1 = 0, 0 \leq x \leq 2 π

10 = 6.7 + 13.75 tan (θ) - 4.718 (1 + tan^{2} (θ))

2 cos (3 θ) = - 3

What is the general solution for 2cos(x)=sin^2(x) ?
The general solution for 2cos(x)=sin^2(x) is x=1.14371…+2pin,x=2pi-1.14371…+2pin