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

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

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Popular >

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

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

cos (x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

2 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)

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

Add the numbers:

1 + 1 = 2

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

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

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

Factor

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

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

Factor out common term

cos (x)

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

cos (x) (- 1 + cos (2 x) - 2 sin^{2} (x)) = 0

Solving each part separately

cos (x) = 0 or - 1 + cos (2 x) - 2 sin^{2} (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

- 1 + cos (2 x) - 2 sin^{2} (x) = 0 : x = πn

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Simplify

- 1 + cos (2 x) + cos (2 x) - 1 : 2 cos (2 x) - 2

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

Add similar elements:

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

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

Group like terms

= 2 cos (2 x) - 1 - 1

Subtract the numbers:

- 1 - 1 = - 2

= 2 cos (2 x) - 2

= 2 cos (2 x) - 2

- 2 + 2 cos (2 x) = 0

Move

2

to the right side

- 2 + 2 cos (2 x) = 0

Add

2

to both sides

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

Simplify

2 cos (2 x) = 2

2 cos (2 x) = 2

Divide both sides by

2

2 cos (2 x) = 2

Divide both sides by

2

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

Simplify

cos (2 x) = 1

cos (2 x) = 1

General solutions for

cos (2 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}

2 x = 0 + 2 πn

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

x = πn

Combine all the solutions

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

Graph

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sqrt(3)=-2cos(θ)

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Frequently Asked Questions (FAQ)

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