#1**+5 **

There are a number of different ways to do a question like this. Here are some ideas.

http://www.thestudentroom.co.uk/showthread.php?t=1586679

ok now I will give it a go.

$$\begin{array}{rlll}

cos^2\theta+sin^2\theta&=&1\\

cos\theta&=&\sqrt{1-sin^2\theta\\\\

3cosx-4sinx &=&1\\

3\sqrt{1-sin^2x}-4sinx &=&1\qquad &\mbox{ }\\

3\sqrt{1-sin^2x} &=&1+4sinx\qquad &\mbox{Square both sides }\\

9\times(1-sin^2x) &=&1+8sinx+16sin^2x\qquad &\mbox{}\\

9-9sin^2x &=&1+8sinx+16sin^2x\qquad &\mbox{}\\

0 &=&-8+8sinx+25sin^2x\qquad &\mbox{}\\

25sin^2x+8sinx-8 &=&0\qquad &\mbox{}\\

Let \;y=sinx&\\

25y^2+8y-8 &=&0\qquad &\mbox{}\\

y&=&\frac{-8\pm \sqrt{64+800}}{50}\\

sinx&=&\frac{-8\pm \sqrt{16*9*6}}{50}\\

sinx&=&\frac{-8\pm 12\sqrt{6}}{50}\\

\end{array}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{864}}}}\right)}{{\mathtt{50}}}}\right)} = -{\mathtt{48.406\: \!856\: \!678\: \!66^{\circ}}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{864}}}}\right)}{{\mathtt{50}}}}\right)} = {\mathtt{25.332\: \!938\: \!613\: \!029^{\circ}}}$$

These answers should be checked by substituting them back into the original equation but I am going to leave that to you. (I'm too tired)

Melody Aug 25, 2014

#1**+5 **

Best Answer

There are a number of different ways to do a question like this. Here are some ideas.

http://www.thestudentroom.co.uk/showthread.php?t=1586679

ok now I will give it a go.

$$\begin{array}{rlll}

cos^2\theta+sin^2\theta&=&1\\

cos\theta&=&\sqrt{1-sin^2\theta\\\\

3cosx-4sinx &=&1\\

3\sqrt{1-sin^2x}-4sinx &=&1\qquad &\mbox{ }\\

3\sqrt{1-sin^2x} &=&1+4sinx\qquad &\mbox{Square both sides }\\

9\times(1-sin^2x) &=&1+8sinx+16sin^2x\qquad &\mbox{}\\

9-9sin^2x &=&1+8sinx+16sin^2x\qquad &\mbox{}\\

0 &=&-8+8sinx+25sin^2x\qquad &\mbox{}\\

25sin^2x+8sinx-8 &=&0\qquad &\mbox{}\\

Let \;y=sinx&\\

25y^2+8y-8 &=&0\qquad &\mbox{}\\

y&=&\frac{-8\pm \sqrt{64+800}}{50}\\

sinx&=&\frac{-8\pm \sqrt{16*9*6}}{50}\\

sinx&=&\frac{-8\pm 12\sqrt{6}}{50}\\

\end{array}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{864}}}}\right)}{{\mathtt{50}}}}\right)} = -{\mathtt{48.406\: \!856\: \!678\: \!66^{\circ}}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{864}}}}\right)}{{\mathtt{50}}}}\right)} = {\mathtt{25.332\: \!938\: \!613\: \!029^{\circ}}}$$

These answers should be checked by substituting them back into the original equation but I am going to leave that to you. (I'm too tired)

Melody Aug 25, 2014