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