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The equation is

$$\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$$

My solution goes like this

$$ \begin{cases} 3+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x) \\ \frac{\sin(x)}{\sqrt 3}+3\cos(x) \ge 0 \end{cases} $$

$$3(\sin^2(x)+\cos^2(x))+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x)$$ $$2\cos^2(x)+\frac{\sin^2(x)}{3}-3\sin^2(x)+\frac{6}{\sqrt 3}\sin(x)\cos(x)=0$$

I multiply by 3 and divide by $\cos^2(x)$:

$$8\tan^2(x)-6\sqrt{3}\tan(x)-6=0$$

Let $t=\tan(x)$, then

$$4t^2-3\sqrt{3}t-3=0$$

$$t_1=\frac{7\sqrt 3}{2}$$

$$t_2=\frac{11\sqrt 3}{4}$$

The solutions for $x$ would be arc-tangents of these values.

But the textbook's answer is

$$\color{green}{x_1=\frac{\pi}{3}+2\pi n; x_2=-\arctan\left(\frac{\sqrt 3}{4}\right)+2\pi n}$$

Where did I make a mistake?


P.S. From the textbook

enter image description here

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    $\begingroup$ Everything was OK until you solved the quadratic at the end ;-) $\endgroup$ – David Aug 28 '15 at 7:18
  • $\begingroup$ @David - Indeed! I'm glad I almost made it. (0: I factored 25 when I should've 5 out of $\sqrt{75}$, getting $25\sqrt{3}$. (0: $\endgroup$ – CopperKettle Aug 28 '15 at 7:36
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You made a mistake in factorising the following equation:

$$ 4t^2 -3\sqrt3 t -3 = 0$$

$$ t = \frac{3\sqrt 3 \pm \sqrt{75}}{8}$$

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$$4t^2-3\sqrt3t-3=0$$

$$\implies t=\dfrac{3\sqrt3\pm\sqrt{(3\sqrt3)^2-4\cdot4(-3)}}{2\cdot4}=\dfrac{3\sqrt3\pm5\sqrt3}8=?$$

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The roots of $4 t^2-3 \sqrt{3} t-3$ are $-\frac{\sqrt{3}}{4}$ and $\sqrt{3}$.

A slightly different approach to this problem could be the following:

\begin{align} 3+4\cos^2(x)-\frac{\sin^2(x)}{3}-\frac{6}{\sqrt 3}\sin(x)\cos(x)-9\cos^2(x)=\frac{1}{3} \left(-3 \sqrt{3} \sin (2 x)-7 \cos (2 x)+1\right) \end{align} With this we have $$3 \sqrt{3} \sin (2 x)+7 \cos (2 x)=2\sqrt{19}\sin(2x+y)$$ where $y=\arcsin\frac{7}{2\sqrt{19}}$. Hence the roots can be found as \begin{align} x&=\frac12\Big(\arcsin\frac{1}{2\sqrt{19}}-\arcsin\frac{7}{2\sqrt{19}}\Big)+2\pi n\\ x&=\arcsin\frac{1}{2\sqrt{19}}+\arcsin\frac{7}{2\sqrt{19}}+2\pi n\\ \end{align} now note that \begin{align} \arcsin\frac{1}{2\sqrt{19}}+\arcsin\frac{7}{2\sqrt{19}}&=\arcsin( \frac{1}{2\sqrt{19}}\sqrt{1-\frac{7}{2\sqrt{19}}^2} + \frac{7}{2\sqrt{19}}\sqrt{1-\frac{1}{2\sqrt{19}}^2})\\ &=\arcsin\frac{\sqrt3}{2}\\ &=\frac{\pi}{3} \end{align} Further using $\tan(\alpha+\beta)$ and $\tan(\arcsin x)=\frac{x}{\sqrt{1-x^2}}$ we obtain $$\arcsin\frac{1}{2\sqrt{19}}-\arcsin\frac{7}{2\sqrt{19}}=-2\arctan \frac{\sqrt3}{4}.$$

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  • $\begingroup$ I'm afraid I don't understand where $\frac{1}{3} \left(-3 \sqrt{3} \sin (2 x)-7 \cos (2 x)+1\right)$ comes from in the third line of your answer. Is this the result of a transformation? I couldn't get a similar result. $\endgroup$ – CopperKettle Aug 28 '15 at 8:42
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    $\begingroup$ I use $\sin 2x = 2 \sin x \cos x$ and $\cos 2 x = 2 \cos ^2 x -1$. I also use $\sin^2 x =1- \cos^2 x$. $\endgroup$ – Math-fun Aug 28 '15 at 9:42
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The solution of the quadratic equation $4t^2 -3\sqrt3 t -3 = 0$ is wrong. Here, $D = b^2 -4ac$ i.e. D=75.
Hence, $t = \frac{3√3 + or - √D}{2*4}$
$t = \frac{3√3 + 5√3}{8}$
And $t = \frac{3√3 -5√3}{8}$
Hence, $ t = √3, -\frac{\sqrt3}{4}$
Now you will get the desired result.

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