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# 3cosx-4sinx=1

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3cosx-4sinx=1

math trigonometry
Guest Aug 25, 2014

### Best Answer

#1
+91469
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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
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### 1+0 Answers

#1
+91469
+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

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