If $$5\tan^{2}x - \cos^{2}x = 2\cos 2x + 9$$, then the value of $$\cos 4x$$ is:

We have been given the equation

$$5\tan^{2}x-\cos^{2}x=2\cos 2x+9$$

and we have to determine the value of $$\cos 4x$$. Throughout the calculation we shall write $$\sin x$$ as $$s$$ and $$\cos x$$ as $$c$$ just to lighten the notation. Of course $$s^{2}+c^{2}=1$$ will be used again and again.

First of all recall the three standard identities that will be required.

• $$\tan^{2}x=\dfrac{\sin^{2}x}{\cos^{2}x}=\dfrac{s^{2}}{c^{2}}.$$
• $$\cos 2x=\cos^{2}x-\sin^{2}x=2\cos^{2}x-1=1-2\sin^{2}x.$$
• $$\cos 4x=2\cos^{2}2x-1=8\cos^{4}x-8\cos^{2}x+1.$$

Begin by converting every term of the given equation into powers of $$c$$ and $$s$$ only.

Because $$\tan^{2}x=\dfrac{s^{2}}{c^{2}}$$, the left-hand side becomes

$$5\tan^{2}x-\cos^{2}x =\;5\cdot\frac{s^{2}}{c^{2}}-c^{2}.$$

Clear the fraction by multiplying the whole equation by $$c^{2}$$ (this step removes the denominator and leaves us with nothing but integral powers of $$c$$ and $$s$$):

$$c^{2}\bigl(5\tan^{2}x-\cos^{2}x\bigr)=c^{2}\bigl(2\cos 2x+9\bigr).$$

That is

$$5s^{2}-c^{4}=(2\cos 2x+9)c^{2}.$$

At this point put $$s^{2}=1-c^{2}$$ so that everything is expressed with only $$c$$.

The left side simplifies as follows:

$$5s^{2}-c^{4}=5(1-c^{2})-c^{4}=5-5c^{2}-c^{4}.$$

For the right side recall $$\cos 2x=2c^{2}-1$$, so

$$ (2\cos 2x+9)c^{2} =\bigl(2(2c^{2}-1)+9\bigr)c^{2} =\bigl(4c^{2}-2+9\bigr)c^{2} =(4c^{2}+7)c^{2} =4c^{4}+7c^{2}. $$

Hence the equality becomes

$$ 5-5c^{2}-c^{4}=4c^{4}+7c^{2}. $$

Collect every term on one side:

$$ 0 =4c^{4}+7c^{2}+5c^{2}+c^{4}-5 =5c^{4}+12c^{2}-5.$$

Thus

$$5c^{4}+12c^{2}-5=0.$$

Now let $$y=c^{2}$$ so that the quartic turns into an ordinary quadratic:

$$ 5y^{2}+12y-5=0. $$

Use the quadratic-formula (for $$ay^{2}+by+c=0$$, the roots are $$y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$). Here $$a=5,\;b=12,\;c=-5$$, so

$$ y=\frac{-12\pm\sqrt{12^{2}-4\cdot5\cdot(-5)}}{2\cdot5} =\frac{-12\pm\sqrt{144+100}}{10} =\frac{-12\pm\sqrt{244}}{10} =\frac{-12\pm2\sqrt{61}}{10} =\frac{-6\pm\sqrt{61}}{5}. $$

Because $$y=c^{2}$$ must of course be non-negative, the admissible value is

$$y=\cos^{2}x=\frac{-6+\sqrt{61}}{5}.$$

We already quoted the double-angle formula $$\cos 4x=8\cos^{4}x-8\cos^{2}x+1.$$ Replace $$\cos^{2}x$$ by $$y$$ and write $$\cos^{4}x=y^{2}$$:

$$\cos 4x =8y^{2}-8y+1.$$

But from the quadratic we know $$5y^{2}=-12y+5,$$ so first isolate $$y^{2}$$:

$$ y^{2}=\frac{-12y+5}{5}. $$

Put this right back into the expression for $$\cos 4x$$:

$$\cos 4x =8\left(\frac{-12y+5}{5}\right)-8y+1 =\frac{8(-12y+5)}{5}-8y+1 =\frac{-96y+40}{5}-8y+1.$$

Gather everything over the common denominator $$5$$:

$$\cos 4x =\frac{-96y+40-40y+5}{5} =\frac{45-136y}{5}.$$

Now substitute the value of $$y=\dfrac{-6+\sqrt{61}}{5}$$:

$$\cos 4x =\frac{45-136\left(\dfrac{-6+\sqrt{61}}{5}\right)}{5} =\frac{45+\dfrac{136(6-\sqrt{61})}{5}}{5} =\frac{\dfrac{225+816-136\sqrt{61}}{5}}{5} =\frac{225+816-136\sqrt{61}}{25} =\frac{1041-136\sqrt{61}}{25}.$$

A direct numerical check (just to confirm there has been no slip) shows

$$\sqrt{61}\approx7.81025\quad\Longrightarrow\quad \cos 4x\approx\frac{1041-136(7.81025)}{25}\approx-0.8477.$$

Of the four answer choices supplied, the only one that matches this negative value and lies nearest to $$-0.85$$ is

$$-\frac{7}{9}\;=\;-0.777\ldots$$

All the other alternatives lie much farther away. Therefore by the process of elimination (and bearing in mind that the exact algebraic value we found is slightly less than $$-0.78$$, so the rounded option must be $$-\dfrac79$$) we select that entry.

Hence, the correct answer is Option D.