0 like 0 dislike
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​
by

2 Answers

0 like 0 dislike
You don't need to single out [tex]\cos \theta[/tex] to solve this question.

To solve:

[tex]\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}[/tex]

So for the given interval [tex]0\leq \theta \leq 2 \pi[/tex]

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

(As confirmed by the attached graph)
by
0 like 0 dislike
[tex]\boxed{\bf cos4\theta=2cos^2(2\theta)-1}[/tex]

We shall use A inplace of theta

[tex]\\ \rm\Rrightarrow 3cos4A=-2[/tex]

[tex]\\ \rm\Rrightarrow cos4A=\dfrac{-2}{3}[/tex]

[tex]\\ \rm\Rrightarrow 2cos^2(2A)=\dfrac{-2}{3}[/tex]

[tex]\\ \rm\Rrightarrow cos^2(2A)=\dfrac{-2}{6}[/tex]

[tex]\\ \rm\Rrightarrow cos^2(2A)=\dfrac{-1}{3}[/tex]

[tex]\\ \rm\Rrightarrow 1-sin^2(2A)=\dfrac{1}{9}[/tex]

[tex]\\ \rm\Rrightarrow 2cos^2A-1=\dfrac{1}{9}[/tex]

[tex]\\ \rm\Rrightarrow 2cos^2A=\dfrac{10}{9}[/tex]

[tex]\\ \rm\Rrightarrow cos^2A=\dfrac{5}{9}[/tex]

[tex]\\ \rm\Rrightarrow cosA=\pm\dfrac{\sqrt{5}}{9}[/tex]

Using calculator

[tex]\\ \rm\Rrightarrow cosA=0.7297+2n\pi[/tex]

And

[tex]\\ \rm\Rrightarrow cosA=2\pi-0.7297+2n\pi[/tex]
by
Welcome to AskTheTask.com, where understudies, educators and math devotees can ask and respond to any number related inquiry. Find support and replies to any numerical statement including variable based math, geometry, calculation, analytics, geometry, divisions, settling articulation, improving on articulations from there, the sky is the limit. Find solutions to numerical problems. Help is consistently 100 percent free!

Questions

No related questions found