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

The general solution for 2cos^2(x)+cos(x)=cos(2x) is x=pi+2pin

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2cos^2(x)+cos(x)=cos(2x)

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

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

cos (2 x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Add similar elements:

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

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

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

Use the Pythagorean identity:

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

= cos (x) + 1

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

cos (x) = - 1

cos (x) = - 1

General solutions for

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

What is the general solution for 2cos^2(x)+cos(x)=cos(2x) ?
The general solution for 2cos^2(x)+cos(x)=cos(2x) is x=pi+2pin