What is the general solution for -4+cos(4θ)=(-8+sqrt(3))/2 ?

The general solution for -4+cos(4θ)=(-8+sqrt(3))/2 is θ= pi/(24)+(pin)/2 ,θ= pi/2-pi/(24)+(pin)/2

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-4+cos(4θ)=(-8+sqrt(3))/2

- 4 + cos (4 θ) = \frac{- 8 + 3}{2}

θ = \frac{π}{24} + \frac{πn}{2}, θ = \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

Solution steps

- 4 + cos (4 θ) = \frac{- 8 + 3}{2}

Move

4

to the right side

cos (4 θ) = \frac{- 8 + 3}{2} + 4

Apply trig inverse properties

4 θ = arccos (\frac{- 8 + 3}{2} + 4) + 2 πn, 4 θ = 2 π - arccos (\frac{- 8 + 3}{2} + 4) + 2 πn

Solve

4 θ = arccos (\frac{- 8 + 3}{2} + 4) + 2 πn : θ = \frac{π}{24} + \frac{πn}{2}

Solve

4 θ = 2 π - arccos (\frac{- 8 + 3}{2} + 4) + 2 πn : θ = \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

θ = \frac{π}{24} + \frac{πn}{2}, θ = \frac{π}{2} - \frac{π}{24} + \frac{πn}{2}

3 cos (x) = 1 + sin (x)

6 cos^{2} (\frac{x}{2}) + cos (x) + 1 = 0

- 2 sin (θ) = 1

cos (θ) = - \frac{7}{3}

0 = - 6 sin (x) + 6 cos (2 x)

What is the general solution for -4+cos(4θ)=(-8+sqrt(3))/2 ?
The general solution for -4+cos(4θ)=(-8+sqrt(3))/2 is θ= pi/(24)+(pin)/2 ,θ= pi/2-pi/(24)+(pin)/2