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3cosx-4sinx=1

math trigonometry
 Aug 25, 2014

Best Answer 

 #1
avatar+109766 
+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)   

 Aug 25, 2014
 #1
avatar+109766 
+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

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