Question

The function f is such that f(x) = (2sinx)^2-(3cosx)^2 for 0<x<pi
i) express f(x) in the form a+b(cos^2)x stating the values of a and b
ii) state the greatest and least values of f(x)
iii) solve the equation f(x) +1=0

Answer 1

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Task. The function $f$ is such that $f(x)=(2\sin x)^{2}-(3\cos x)^{2}$ for $0\leq x\leq\pi$.

i) express $f(x)$ in the form $a+b(\cos x)^{2}$ stating the values of $a$ and $b$;

ii) state the greatest and least values of $f(x)$

iii) solve the equation $f(x)+1=0$

Solution.

i) Recall that for any $x$ we have that

$(\sin x)^{2}+(\cos x)^{2}=1.$

Therefore

$(\sin x)^{2}=1-(\cos x)^{2},$

whence

$f(x)$ $=(2\sin x)^{2}-(3\cos x)^{2}=4(\sin x)^{2}-9(\cos x)^{2}$

$=4(1-(\cos x)^{2})-9(\cos x)^{2}$

$=4-4(\cos x)^{2}-9(\cos x)^{2}=4-13(\cos x)^{2},$

so

$f(x)=a+b(\cos x)^{2},$

where

$a=4,\qquad b=-13.$

ii) Notice that the maximal value of $(\cos x)^{2}$ is $1$, whence the maximal value of $f$ is

$\max\,f=4-13\cdot 1=-9.$

iii) Let us solve the equation

$f(x)+1=0,$

that is

$4-13(\cos x)^{2}+1=0,$

$13(\cos x)^{2}=5$

$(\cos x)^{2}=5/13,$

$\cos x=\pm\sqrt{5/13}.$

Since $x\in[0,\pi]$, it follows that

$\cos x=\sqrt{5/13}\qquad\Rightarrow\qquad x=\arccos\sqrt{5/13},$

$\cos x=-\sqrt{5/13}\qquad\Rightarrow\qquad x=\pi-\arccos\sqrt{5/13}.$

So we have two solutions:

$\arccos\sqrt{5/13},\qquad\pi-\arccos\sqrt{5/13}.$