What is the general solution for solvefor t,x=cos(t)+cos(2t) ?

The general solution for solvefor t,x=cos(t)+cos(2t) is

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solvefor t,x=cos(t)+cos(2t)

solve for

t, x = cos (t) + cos (2 t)

t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

Solution steps

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

Switch sides

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

Subtract

x

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (t) = \frac{- 1 + 8 x + 9}{4}, cos (t) = \frac{- 1 - 8 x + 9}{4}

cos (t) = \frac{- 1 + 8 x + 9}{4} : t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn

cos (t) = \frac{- 1 - 8 x + 9}{4} : t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

Combine all the solutions

t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

x, sin (x) = - 1

9 - 9 sin (θ) = 6 cos^{2} (θ)

cot (2 θ) = 0

sec (2 x) = - 2

cos (x) = \frac{5}{7}

What is the general solution for solvefor t,x=cos(t)+cos(2t) ?
The general solution for solvefor t,x=cos(t)+cos(2t) is