Question

Find all exact solutions
2cos^2x=sinx+1
Please show work thank you!

Answer 1

`\displaystyle x=\left\{(\pi)/(6)+2n\pi, (5\pi)/(6)+2n\pi, (3\pi)/(2)+2n\pi\right\}, n\in\mathbb{Z}`

Explanation:

We are given the equation:

`2\cos^2(x)=\sin(x)+1`

And we want to find all solutions for x.

First, we should put the equation into terms of only one trigonometric ratio.

Since we are given cos²(x), we can turn this into sine. Recall the Pythagorean Identity which states:

`\sin^2(x)+\cos^2(x)=1`

Therefore:

`\cos^2(x)=1-\sin^2(x)`

By substitution:

`2(1-\sin^2(x))=\sin(x)+1`

Distribute:

`2-2\sin^2(x)=\sin(x)+1`

Isolate the equation:

`2\sin^2(x)+\sin(x)-1=0`

We can factor:

`(2\sin(x)-1)(\sin(x)+1)=0`

Zero Product Property:

`2\sin(x)-1=0\text{ or } \sin(x)+1=0`

Solve for each case:

`\displaystyle \sin(x)=(1)/(2)\text{ or } \sin(x)=-1`

We can use the unit circle.

sin(x) = 1/2 for every π/6 and 5π/6. So, it will continue every 2π.

sin(x) = -1 every 3π/2. And this will also continue every 2π.

Hence, our solutions are:

`\displaystyle x=\left\{(\pi)/(6)+2n\pi, (5\pi)/(6)+2n\pi, (3\pi)/(2)+2n\pi\right\}, n\in\mathbb{Z}`

Note:

If you only need the solutions within the interval [0, 2π), then it is:

`\displaystyle x=\left\{(\pi)/(6), (5\pi)/(6), (3\pi)/(2)\right\}`