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

The general solution for cos(4θ)=-2/3 is θ=(2.30052…)/4+(pin)/2 ,θ=-(2.30052…)/4+(pin)/2

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cos(4θ)=-2/3

cos (4 θ) = - \frac{2}{3}

θ = \frac{2.30052 \dots}{4} + \frac{πn}{2}, θ = - \frac{2.30052 \dots}{4} + \frac{πn}{2}

Solution steps

cos (4 θ) = - \frac{2}{3}

Apply trig inverse properties

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

Solve

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

Solve

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

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

Show solutions in decimal form

θ = \frac{2.30052 \dots}{4} + \frac{πn}{2}, θ = - \frac{2.30052 \dots}{4} + \frac{πn}{2}

3 + 4 cos (2 x) + 2 cos (4 x) = 0

cos (θ) = - \frac{4}{5}, sin (2 θ), 18 0^{\circ} < θ < 27 0^{\circ}

2 sin (θ) + 1 = 2

2 - 2 sin^{2} (3 x) = 1

4 cos^{2} (x) = 5 - 4 sin (x), 0 \leq x \leq 2 π

What is the general solution for cos(4θ)=-2/3 ?
The general solution for cos(4θ)=-2/3 is θ=(2.30052…)/4+(pin)/2 ,θ=-(2.30052…)/4+(pin)/2