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

Solution

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

Solution

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

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

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 (2 x) cos (x) + 2 sin (2 x) sin (x)

Use the Double Angle identity:

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

= 1 + 2 cos (2 x) cos (x) + 2 \cdot 2 sin (x) cos (x) sin (x)

2 \cdot 2 sin (x) cos (x) sin (x) = 4 sin^{2} (x) cos (x)

2 \cdot 2 sin (x) cos (x) sin (x)

Multiply the numbers:

2 \cdot 2 = 4

= 4 sin (x) cos (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= 4 cos (x) sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 4 cos (x) sin^{2} (x)

= 1 + 2 cos (2 x) cos (x) + 4 sin^{2} (x) cos (x)

Use the Double Angle identity:

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

= 1 + 2 (1 - 2 sin^{2} (x)) cos (x) + 4 sin^{2} (x) cos (x)

Simplify

1 + 2 (1 - 2 sin^{2} (x)) cos (x) + 4 sin^{2} (x) cos (x) : 1 + 2 cos (x)

1 + 2 (1 - 2 sin^{2} (x)) cos (x) + 4 sin^{2} (x) cos (x)

= 1 + 2 cos (x) (1 - 2 sin^{2} (x)) + 4 sin^{2} (x) cos (x)

Expand

2 cos (x) (1 - 2 sin^{2} (x)) : 2 cos (x) - 4 sin^{2} (x) cos (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2 cos (x), b = 1, c = 2 sin^{2} (x)

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

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

Simplify

2 \cdot 1 \cdot cos (x) - 2 \cdot 2 sin^{2} (x) cos (x) : 2 cos (x) - 4 sin^{2} (x) cos (x)

2 \cdot 1 \cdot cos (x) - 2 \cdot 2 sin^{2} (x) cos (x)

Multiply the numbers:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= 2 cos (x) - 4 sin^{2} (x) cos (x)

= 2 cos (x) - 4 sin^{2} (x) cos (x)

= 1 + 2 cos (x) - 4 sin^{2} (x) cos (x) + 4 sin^{2} (x) cos (x)

Add similar elements:

- 4 sin^{2} (x) cos (x) + 4 sin^{2} (x) cos (x) = 0

= 1 + 2 cos (x)

= 1 + 2 cos (x)

1 + 2 cos (x) = 0

Move

1

to the right side

1 + 2 cos (x) = 0

Subtract

1

from both sides

1 + 2 cos (x) - 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

Popular Examples

5cos(3x)=4

5 cos (3 x) = 4

tan(θ/2+pi/4)=-1

tan (\frac{θ}{2} + \frac{π}{4}) = - 1

2cos^2(θ)-11cos(θ)+5=0

2 cos^{2} (θ) - 11 cos (θ) + 5 = 0

4sin(2x)-1=0

4 sin (2 x) - 1 = 0

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

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

Frequently Asked Questions (FAQ)

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