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Solve the equation in the interval 0 to 2pi.

3 Cos 4θ = -2

I keep trying to single out Cos θ, but once I get to 4θ I'm stuck, the closest to an answer I got was -1/6 but I dont know what that would be on the interval 0 to 2pi​

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You don't need to single out $$\cos \theta$$ to solve this question.

To solve:

\begin{aligned}3 \cos 4 \theta & = -2\\\\\cos 4 \theta & = -\dfrac{2}{3}\\\\4 \theta & = \cos^{-1}\left( -\dfrac{2}{3\right)}\\\\\implies 4 \theta & =2.30053... \pm 2 \pi n, -2.30053...\pm 2 \pi n\\\\\theta & =\dfrac{2.30053...}{4} \pm \dfrac{\pi n}{2}, -\dfrac{2.30053...}{4}\pm \dfrac{\pi n}{2}\end{aligned}

So for the given interval $$0\leq \theta \leq 2 \pi$$

$$\implies \theta =0.575, 2.146, 3.717, 5.288, 0.996, 2.566, 4.137, 5.708\:\:(\sf 3 \: d.p.)$$

(As confirmed by the attached graph)
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$$\boxed{\bf cos4\theta=2cos^2(2\theta)-1}$$

We shall use A inplace of theta

$$\\ \rm\Rrightarrow 3cos4A=-2$$

$$\\ \rm\Rrightarrow cos4A=\dfrac{-2}{3}$$

$$\\ \rm\Rrightarrow 2cos^2(2A)=\dfrac{-2}{3}$$

$$\\ \rm\Rrightarrow cos^2(2A)=\dfrac{-2}{6}$$

$$\\ \rm\Rrightarrow cos^2(2A)=\dfrac{-1}{3}$$

$$\\ \rm\Rrightarrow 1-sin^2(2A)=\dfrac{1}{9}$$

$$\\ \rm\Rrightarrow 2cos^2A-1=\dfrac{1}{9}$$

$$\\ \rm\Rrightarrow 2cos^2A=\dfrac{10}{9}$$

$$\\ \rm\Rrightarrow cos^2A=\dfrac{5}{9}$$

$$\\ \rm\Rrightarrow cosA=\pm\dfrac{\sqrt{5}}{9}$$

Using calculator

$$\\ \rm\Rrightarrow cosA=0.7297+2n\pi$$

And

$$\\ \rm\Rrightarrow cosA=2\pi-0.7297+2n\pi$$
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