Question

Hello, can you please help me solve this question ASAP!!!

Answer 1

Step 1:

In this question, we have that:

Step 2:

Part A:

We are meant to show that the equation:

`5sinx=1+2cos^2x`

can be written in the form

`2sin^2\text{x + 5 sin x - 3=0}`

Proof:

`\begin{gathered} \text{5 sin x = 1 + 2 cos }^2x\text{ } \\ \text{But cos}^2x+sin^2x\text{ = 1} \\ \text{Then,} \\ \cos ^2x=1-sin^2x\text{ } \\ \text{Hence,} \\ 5sinx=1+2(1-sin^2x_{}) \\ 5sinx=1+2-2sin^2x \\ 5sinx=3-2sin^2x \end{gathered}`

Re-arranging, we have that:

`2sin^2x\text{ + 5 sin x - 3 = 0 }`

Part B:

b) Hence, solve for x in the interval:

`0\text{ }\leq\text{ x }\leq\text{ 2}\pi`