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

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

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

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

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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= 2 cos (x) - 1

2 cos (x) - 1 = 0

Move

1

to the right side

2 cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

2 cos (x) = 1

2 cos (x) = 1

Divide both sides by

2

2 cos (x) = 1

Divide both sides by

2

\frac{2 cos ( x )}{2} = \frac{1}{2}

Simplify

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

2 cos (x) - 1 = 0, 0 \leq x < 2 π

cos (t) = \frac{3}{5}

cos (4 θ) = - \frac{2}{3}

3 + 4 cos (2 x) + 2 cos (4 x) = 0

cos (θ) = - \frac{4}{5}, sin (2 θ), 18 0^{\circ} < θ < 27 0^{\circ}

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