I had the following equation to do:

7 sinx + 4 cosx = 8.

My solutions (to 3 decimal places) were 0.927 and 1.106.

Can someone verify these please? If they are not correct, could someone show me how to do it?

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- Oct 31st 2007, 01:39 PMty2391An Equation
I had the following equation to do:

7 sinx + 4 cosx = 8.

My solutions (to 3 decimal places) were 0.927 and 1.106.

Can someone verify these please? If they are not correct, could someone show me how to do it? - Oct 31st 2007, 01:50 PMJhevon
- Oct 31st 2007, 01:53 PMCaptainBlack
- Oct 31st 2007, 02:26 PMty2391
Yup I got 1.176, just a typo.

The way I did it was representing 7 sinx + 4 cosx as a single sine function.

Once I did that, I just equated that to 8, and solved for x. - Oct 31st 2007, 04:59 PMThePerfectHacker
$\displaystyle 7\sin x + 4\cos x = 8$

$\displaystyle \sqrt{4^2+7^2} \left( \frac{7}{\sqrt{4^2+7^2}}\sin x + \frac{4}{\sqrt{4^2+7^2}}\cos x \right) = 8$

Let $\displaystyle \theta = \tan^{-1} \frac{4}{7}$ then,

$\displaystyle \sqrt{4^2+7^2} (\sin x \cos \theta + \cos x \sin \theta) = 8$

Thus,

$\displaystyle \sin (x + \theta) = \frac{8}{\sqrt{55}}$

Now you can solve for $\displaystyle x$. - Oct 31st 2007, 05:53 PMJhevon
hmm. what i did was:

$\displaystyle 7 \sin x + 4 \cos x = 8$

$\displaystyle \Rightarrow 7 \sin x = 8 - 4 \cos x$

$\displaystyle \Rightarrow 49 \sin^2 x = ( 8 - 4 \cos x)^2 $

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i think by now you see where i am going with this. simplifying what i have above leads to a quadratic equation in cosine, and dealing with quadratics is easy

interesting solution