Show that the equation $\displaystyle 15 cos x = 13+sin x$ may be written as a quadratic equation in $\displaystyle sin x$.
Hence solve the equation, giving all values of "x" such that 0<=x<=360.
Didn't help me, this is what I did:
$\displaystyle 15 cos^2x-(13+sin x)^2$
$\displaystyle 225cos^4x=(13+sin x)(13+sin x)$
$\displaystyle 225cos^4x=196+26sin x +sin^2x$
Was that actually going good and I simply didn't notice or did I get what you meant with square both sides wrong?
I am not sure what you are doing in the above. How did squaring cosine produce $\displaystyle \cos^4x?$ Is the equation in the original post the correct equation?
You should have
$\displaystyle 15\cos x=13+\sin x$
$\displaystyle \Rightarrow225\cos^2x=\sin^2x+26\sin x+169$
$\displaystyle \Rightarrow225\left(1-\sin^2x\right)=\sin^2x+26\sin x+169$
$\displaystyle \Rightarrow225-225\sin^2x=\sin^2x+26\sin x+169$
$\displaystyle \Rightarrow-226\sin^2x-26\sin x+56=0.$
There is your quadratic, now solve it.
Yes, Moo. Why the red things?