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Popular >

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

Solution

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

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{4} + πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Factor

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

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

Factor out common term

cos (x)

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

Apply exponent rule:

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

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

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

Factor out common term

sin (x)

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

Factor out common term

(1 - 2 sin (x))

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

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

Solving each part separately

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

1 - 2 sin (x) = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

1 - 2 sin (x) = 0

Move

1

to the right side

1 - 2 sin (x) = 0

Subtract

1

from both sides

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

Simplify

- 2 sin (x) = - 1

- 2 sin (x) = - 1

Divide both sides by

- 2

- 2 sin (x) = - 1

Divide both sides by

- 2

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

Simplify

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

cos (x) + sin (x) = 0 : x = \frac{3 π}{4} + πn

cos (x) + sin (x) = 0

Rewrite using trig identities

cos (x) + sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{cos ( x ) + sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

1 + \frac{sin ( x )}{cos ( x )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 + tan (x) = 0

1 + tan (x) = 0

Move

1

to the right side

1 + tan (x) = 0

Subtract

1

from both sides

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

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{4} + πn

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