Question

Use inverse functions where needed to find all solutions of the equation in the interval [0,2pi]

Answer 1

Step 1

Solve the quadratic equation for sin x

`\begin{gathered} \text{let sinx = m} \\ 2m^2-19m+9=0 \end{gathered}`
`\begin{gathered} The\text{ factors to simplify the quadratic equation are} \\ -18m\text{ and -m} \end{gathered}`
`\begin{gathered} \text{Hence,} \\ 2m^2-18m-m+9=0 \\ (2m^2-18m)(-m+9)=0 \\ 2m(m-9)-1(m-9)=0_{}_{} \end{gathered}`
`\begin{gathered} (2m-1)(m-9)\text{ = 0} \\ (2m-1)\text{ = 0} \\ \text{or} \\ m-9=0 \\ 2m=1 \\ or \\ m=\text{ 9} \\ \text{Hence,} \\ m=(1)/(2)or\text{ 9} \end{gathered}`

Step 2

But m = sin x

`\begin{gathered} \text{Therefore,} \\ \sin x=9 \\ x=\sin ^(-1)(9)\text{ = No solution} \\ \sin x=(1)/(2) \\ x=\sin ^(-1)((1)/(2)) \\ x=30^(\circ) \end{gathered}`