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

The general solution for 2cos(2x)cos(x)+2sin(2x)sin(x)=-1 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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

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

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Solution steps

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

Subtract

- 1

from both sides

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

Rewrite using trig identities

1 + 2 cos (x) = 0

Move

1

to the right side

2 cos (x) = - 1

Divide both sides by

2

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

General solutions for

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

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

5 cos (3 x) = 4

tan (\frac{θ}{2} + \frac{π}{4}) = - 1

2 cos^{2} (θ) - 11 cos (θ) + 5 = 0

4 sin (2 x) - 1 = 0

cos (2 x) - 3 sin (2 x) = 0

What is the general solution for 2cos(2x)cos(x)+2sin(2x)sin(x)=-1 ?
The general solution for 2cos(2x)cos(x)+2sin(2x)sin(x)=-1 is x=(2pi)/3+2pin,x=(4pi)/3+2pin