<s>[INST] Maximum of a trigonometric expression Maximize $f(x)=4\sin x+48\sin x\cos x+3\cos x+14\sin^2x$ I broke it in 2 parts but then realized they don't have their maxima at the same points. So I am stuck. This is what I did. I wrote it as $$3\cos x+4\sin x+2\sin x(24\cos x+7\sin x)$$ Now we can maximize the first two terms and the terms inside the bracket.
 [/INST] $f(x)=4\sin x+48\sin x\cos x+3\cos x+14\sin^2x$
Step 1: Take derivative and equate it to zero
$f'(x)=4\cos x+48(\cos^x-\sin^2 x)-3\sin x+14(2\sin x\cos x)$
$\implies 4\cos x + 48\cos 2x-3\sin x+14\sin 2x=0$
Step 2: Find the solution to above equation. 
Step 3: It will attain a maxima if $f''(x)<0$, that means second derivative of the function.
Follow this wolfram Alpha link for solutions.
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