SOLUTION:
Step 1:
In this question, we have that:
Step 2:
Part A:
We are meant to show that the equation:
[tex]5sinx=1+2cos^2x[/tex]
can be written in the form
[tex]2sin^2\text{x + 5 sin x - 3=0}[/tex]
Proof:
[tex]\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}[/tex]
Re-arranging, we have that:
[tex]2sin^2x\text{ + 5 sin x - 3 = 0 }[/tex]
Part B:
b) Hence, solve for x in the interval:
[tex]0\text{ }\leq\text{ x }\leq\text{ 2}\pi[/tex]
