Math Help - trig problem

1. trig problem

This is on a C2 exam past paper and I haven't studied this but I want to know how to do it:

It is a two part question:

a] Find all values of $\theta$ in the range $0^{\circ} \leq \theta \leq 360^{\circ}$ satisfying $12\ sin^2\ \theta - 5\ cos\ \theta = 9$

b] Find all values of $x$ in the range $0^{\circ} \leq x \leq 180^{\circ}$ satisfying $sin\ (3x+15^{\circ})\ =\ 0.5$

Can someone please go through step by step or as much as they can to try and help me?

2. Hi

a] Replace $\ sin^2 \theta$ by $1 - \cos^2\theta$ and you will get a second degree equation in $\cos \theta$

b] $\sin (3x+15^{\circ}) = 0.5 = \sin 30^{\circ}$
$\sin a = \sin b$ leads to $a = b + 360^{\circ}k$ or $a = 180^{\circ} - b + 360^{\circ}k$ with k real

3. Hello, db5vry!

(a) Solve: . $12\sin^2\!\theta - 5\cos\theta \:=\:0,\,\text{ for }0^o \leq \theta \leq 360^o$

We have: . $12(1-\cos^2\!\theta) - 5\cos\theta \:=\:0 \quad\Rightarrow\quad 12\cos^2\!\theta + 5\cos\theta - 3 \:=\:0$

Factor: . $(3\cos\theta-1)(4\cos\theta + 3) \:=\:0$

And we have:

. . $\begin{array}{ccccc}3\cos\theta -1\:=\:0 & \Rightarrow & \cos\theta \:=\:\tfrac{1}{3} &\Rightarrow&\theta \:\approx\:70.5^o,\:289.5^o \\ \\

4\cos\theta + 3 \:=\:0 &\Rightarrow& \cos\theta \:=\:\text{-}\tfrac{3}{4} & \Rightarrow & \theta \:\approx\:138.6^o,\:221.4^o\end{array}$

(b) Solve: . $\sin(3x+15^o) \:=\:\tfrac{1}{2},\;\text{ for }0^o \leq x \leq 180^o$

We have: . $\sin(3x+15^o) \:=\:\tfrac{1}{2}$

. . $\begin{array}{cc}\text{Then:} & 3x + 15^o \:=\:\left\{30^o,\:150^o,\:390^o,\:510^o\right\} \\ \\

\text{And:} & 3x \:=\:\left\{15^o,\:135^o,\:375^o,\:495^o\right\} \\ \\

\text{Hence:} & x \:=\:\left\{5^o,\:45^o,\:125^o,\:165^o\right\}\end {array}$