Question

7.2. I have a question about advanced trig equations that I really need help with picture included

Answer 1

`\theta=(\pi)/(6),\: \theta=(11\pi)/(6)`

1) Let's start out isolating the cosine by dividing both sides by 2

`\begin{gathered} 2\cos \mleft(\theta\mright)=√(3) \\ (2\cos\left(θ\right))/(2)=(√(3))/(2) \\ \cos \mleft(\theta\mright)=(√(3))/(2) \\ \end{gathered}`

2) From that we can find two general solutions in which the cosine of theta yields the square root of 3 over two:

`\begin{gathered} \cos (30^(\circ))or\cos ((\pi)/(6))\text{ and }cos(330^(\circ)or(11)/(6)\pi)=\frac{\sqrt[]{3}}{2} \\ \theta=(\pi)/(6)+2\pi n,\: \theta=(11\pi)/(6)+2\pi n \end{gathered}`

But not that there is a restraint, so we can write out the solution as:

`\theta=(\pi)/(6),\: \theta=(11\pi)/(6)`