+0

# 3cosx-4sinx=1

0
1320
1

3cosx-4sinx=1

math trigonometry
Aug 25, 2014

### Best Answer

#1
+97500
+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.

$$\begin{array}{rlll} cos^2\theta+sin^2\theta&=&1\\ cos\theta&=&\sqrt{1-sin^2\theta\\\\ 3cosx-4sinx &=&1\\ 3\sqrt{1-sin^2x}-4sinx &=&1\qquad &\mbox{ }\\ 3\sqrt{1-sin^2x} &=&1+4sinx\qquad &\mbox{Square both sides }\\ 9\times(1-sin^2x) &=&1+8sinx+16sin^2x\qquad &\mbox{}\\ 9-9sin^2x &=&1+8sinx+16sin^2x\qquad &\mbox{}\\ 0 &=&-8+8sinx+25sin^2x\qquad &\mbox{}\\ 25sin^2x+8sinx-8 &=&0\qquad &\mbox{}\\ Let \;y=sinx&\\ 25y^2+8y-8 &=&0\qquad &\mbox{}\\ y&=&\frac{-8\pm \sqrt{64+800}}{50}\\ sinx&=&\frac{-8\pm \sqrt{16*9*6}}{50}\\ sinx&=&\frac{-8\pm 12\sqrt{6}}{50}\\ \end{array}$$

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

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

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+0 Answers

#1
+97500
+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.

$$\begin{array}{rlll} cos^2\theta+sin^2\theta&=&1\\ cos\theta&=&\sqrt{1-sin^2\theta\\\\ 3cosx-4sinx &=&1\\ 3\sqrt{1-sin^2x}-4sinx &=&1\qquad &\mbox{ }\\ 3\sqrt{1-sin^2x} &=&1+4sinx\qquad &\mbox{Square both sides }\\ 9\times(1-sin^2x) &=&1+8sinx+16sin^2x\qquad &\mbox{}\\ 9-9sin^2x &=&1+8sinx+16sin^2x\qquad &\mbox{}\\ 0 &=&-8+8sinx+25sin^2x\qquad &\mbox{}\\ 25sin^2x+8sinx-8 &=&0\qquad &\mbox{}\\ Let \;y=sinx&\\ 25y^2+8y-8 &=&0\qquad &\mbox{}\\ y&=&\frac{-8\pm \sqrt{64+800}}{50}\\ sinx&=&\frac{-8\pm \sqrt{16*9*6}}{50}\\ sinx&=&\frac{-8\pm 12\sqrt{6}}{50}\\ \end{array}$$

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

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

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

### New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.