Question

Find all solutions to 2sin(theta) = 1 on the interval 0 <= theta < 2pi

Answer 1

Let's determine the solutions of the following:

`\text{ 2 }\cdot\text{ Sin \lparen}\theta)\text{ = 1}`

We get,

`\text{ 2 }\cdot\text{ Sin \lparen}\theta)\text{ = 1}`
`\text{ Sin \lparen}\theta)\text{ = }(1)/(2)`
`\text{ }\theta\text{ = Sin}^(-1)((1)/(2))`
`\text{ }\theta\text{ = }(\pi)/(6)`

Since we are asked for the solution at 0 < Θ < 2π interval, we will be looking at the solution in the 1st and second quadrants.

For the second solution, subtract the reference angle from π to find the solution in the second quadrant.

We get,

`\text{ }\pi\text{ - }(\pi)/(6)\text{ = }(6\pi)/(6)\text{ - }(\pi)/(6)`
`\text{ = }(5\pi)/(6)`

Therefore, the solutions are:

`\text{ }(\pi)/(6),\text{ }(5\pi)/(6)`