# An Equation

• Oct 31st 2007, 02:39 PM
ty2391
An 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, 02:50 PM
Jhevon
Quote:

Originally Posted by ty2391
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?

i got the first answer, but not the second. what did you do? what range are we solving for by the way?
• Oct 31st 2007, 02:53 PM
CaptainBlack
Quote:

Originally Posted by ty2391
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?

The second should be more like 1.176

RonL
• Oct 31st 2007, 03:26 PM
ty2391
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, 05:59 PM
ThePerfectHacker
$7\sin x + 4\cos x = 8$

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

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

$\sqrt{4^2+7^2} (\sin x \cos \theta + \cos x \sin \theta) = 8$
Thus,
$\sin (x + \theta) = \frac{8}{\sqrt{55}}$

Now you can solve for $x$.
• Oct 31st 2007, 06:53 PM
Jhevon
Quote:

Originally Posted by ty2391
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.

hmm. what i did was:

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

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

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

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

Quote:

Originally Posted by ThePerfectHacker
$7\sin x + 4\cos x = 8$

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

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

$\sqrt{4^2+7^2} (\sin x \cos \theta + \cos x \sin \theta) = 8$
Thus,
$\sin (x + \theta) = \frac{8}{\sqrt{55}}$

Now you can solve for $x$.

interesting solution