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The range of $$f(x)=3\cos^2x-8\sqrt3 \cos x\cdot\sin x+5\sin^2x-7$$is given by:
(1)$[-7,7]$
(2)$[-10,4]$
(3)$[-4,4]$
(4)$[-10,7]$

ANS: (2)

My Solution

The equation can be written as: $$3\cos^2x-8\sqrt3 \cos x\cdot \sin x+16\sin^2x-11\sin^2x-7 \\\implies (\sqrt3\cos x-4\sin x)^2-11\sin^2x-7$$

So let $y=\sqrt3\cos x+4\sin x$

$$-\sqrt{(\sqrt3)^2+4^2} \le y\le \sqrt{(\sqrt3)^2+4^2} \implies 19 \le y^2 \le 19 \implies y^2\in[0,19]$$

CASE 1: When $y^2=0$ $$f(x)=0-11\sin^2x-7 \text{ ,taking } \sin^2 x=0\text{, minimum value of }f(x)= -7$$

CASE 2: When $y^2=19$ $$f(x)=19-11\sin^2x-7 \text{ ,taking } \sin^2 x=1\text{, minimum value of }f(x)= 1$$

So my range is $y\in[-7, 1]$

where is the problem?

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    $\begingroup$ Convert this function into form $$a\cos 2x+b\sin 2x+c.$$ $\endgroup$ – Bumblebee Jun 10 '15 at 5:22
  • $\begingroup$ got the equation $4\sqrt3sin2x -cos2x-7$ and finally git the solution. Thanks. $\endgroup$ – anni Jun 10 '15 at 5:31
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    $\begingroup$ Your square in the first line should read $(\sqrt{3}\cos x - 4\sin x)^2$ $\endgroup$ – OnceUponACrinoid Jun 10 '15 at 5:33
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Let's do this. \begin{align} f(x)&=3\cos^2 x−8\sqrt3 \cos x\sin x + 5\sin^2x−7 =\\ &= 3(\cos^2x + \sin^2 x) - 4\sqrt3\cdot 2\sin x\cos x + 2\sin^2 x - 7=\\ &=3 - 4\sqrt 3\sin 2x + 2\sin^2 x - 7 = \\ &=-4 -4\sqrt 3\sin 2x + 1 - \cos 2x=\\ &=-\cos2x -4\sqrt 3\sin 2x-3. \end{align} Now we have $$1\cdot\cos2x +4\sqrt 3\cdot\sin 2x = \sqrt{1^2 + (4\sqrt 3)^2}\sin(2x+\varphi)=7\sin(2x+\varphi),$$ where $\sin\varphi=1/7$, $\cos\varphi=4\sqrt3/7$. So you have $$f(x)=-7\sin(2x+\varphi)-3,$$ and range is $[-7-3,7-3] = [-10, 4]$.

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The problem is when you say $y=0$ you fix the value of $sinx$ as $\sqrt3cosx-4sinx=0$ so $tanx=\frac{\sqrt3}{4}$ and $sin^2x= \frac{3}{19}$ not 0 or 1 . You should write it as the form of $asin2x+bcos2x+c$ as stated in the comment.

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First use the double angle formulas to lower the degree

$$3\frac{\cos(2x)+1}2-\frac82\sqrt3 \sin(2x)+5\frac{1-\cos(2x)}2-7 =-\cos(2x)-4\sqrt3 \sin(2x)-3.$$ The dot product $$(\cos(2x),\sin(2x))\cdot(-1,-4\sqrt3)$$ equals $$1\cdot\sqrt{(-1)^2+(-4\sqrt3)^2}\cdot\cos(\phi)$$ where $\phi$ is the angle between the vectors, hence the range is $$[-3-7,-3+7].$$

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