Question

Solve `11sin(theta)-2=2` for a value of `theta` when `0 <= theta <= pi/2`.?

I think I know how to do the first one but I'm unsure about how to do the second problem

Answer 1

`theta = 0.659`

Explanation:

`5cos2theta+7=cos2theta+8`

Simplifies to (simply by rearranging):
`4cos2theta=1`

We know `theta=1.32` by using inverse cos, and now we need to find out `2theta`.

`2theta=1.32`
`theta = 0.659`

Answer 2

Given: `5cos(2theta)+7=cos(2theta)+8; 0 <= theta <= pi/2`

Subtract 7 from both sides:

`5cos(2theta)=cos(2theta)+1; 0 <= theta <= pi/2`

Subtract `cos(2theta)` from both sides:

`4cos(2theta)=1; 0 <= theta <= pi/2`

Divide both sides by 4:

`cos(2theta)=1/4; 0 <= theta <= pi/2`

Use the inverse cosine on both sides:

`2theta= cos^-1(1/4); 0 <= theta <= pi/2`

Divide both sides by 2:

`theta = 1/2cos^-1(1/4); 0 <= theta <= pi/2`

`theta = 0.659`