Question

Solve the equation on the interval [0, 2\small \pi). Show all work. Do not use a calculator - use your unit circle!

Answer 1

Write out the equation given

`\cos ^2x+2\cos x-3=0`

Let

`\text{Cosx}=P`

Then by substitution, we obtain the equation

`p^2+2p-3=0`

Solve the quadractic equation using factor method

`\begin{gathered} p^2+3p-p-3=0 \\ p(p+3)-1(p+3)=0 \\ (p-1)(p+3)=0 \end{gathered}`

Then we have

`\begin{gathered} p-1=0,p+3=0 \\ \text{Then} \\ p=1,p=-3 \end{gathered}`

Recall that

`\cos x=p`

Hence

`\begin{gathered} \text{when p=1} \\ \cos x=1 \\ \text{Then } \\ x=\cos ^(-1)(1)=0 \\ \text{hence } \\ x=0 \end{gathered}`

Similarly,

`\begin{gathered} \text{When p=-3} \\ \cos x=-3 \\ x=\cos ^(-1)(-3) \\ x=no\text{ solution} \end{gathered}`

Therefore x=0 is the only valid solution on the given interval [0,2π).

Answer; x=0