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

The general solution for 2cos(x)cos(2x)-2sin(x)sin(2x)=-1 is x=(2pi)/9+(2pin)/3 ,x=(4pi)/9+(2pin)/3

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

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

x = \frac{2 π}{9} + \frac{2 πn}{3}, x = \frac{4 π}{9} + \frac{2 πn}{3}

Solution steps

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

Subtract

- 1

from both sides

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

Rewrite using trig identities

1 + 2 cos (3 x) = 0

Move

1

to the right side

2 cos (3 x) = - 1

Divide both sides by

2

cos (3 x) = - \frac{1}{2}

General solutions for

cos (3 x) = - \frac{1}{2}

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

Solve

3 x = \frac{2 π}{3} + 2 πn : x = \frac{2 π}{9} + \frac{2 πn}{3}

Solve

3 x = \frac{4 π}{3} + 2 πn : x = \frac{4 π}{9} + \frac{2 πn}{3}

x = \frac{2 π}{9} + \frac{2 πn}{3}, x = \frac{4 π}{9} + \frac{2 πn}{3}

- 19 sin (θ) + \frac{19}{2} = 0

y, e^{8 x} = sin (x + 3 y)

π cos (π x) + π sin (π x) = 0

144 = 100 + 6.9 3^{2} - 2 \cdot 10 \cdot 6.93 \cdot cos (x)

0.09 = 0.18 sin (1.082 t)

What is the general solution for 2cos(x)cos(2x)-2sin(x)sin(2x)=-1 ?
The general solution for 2cos(x)cos(2x)-2sin(x)sin(2x)=-1 is x=(2pi)/9+(2pin)/3 ,x=(4pi)/9+(2pin)/3