   Question Answers

# The number of solutions in $x \in [0,2\pi ]$ for which $|\sqrt {2{{\sin }^4}x + 18{{\cos }^2}x} - \sqrt {2{{\cos }^4}x + 18{{\sin }^2}x} | = 1$ isA) 4B) 2C) 6D) 8

Hint: This is a particular problem of trigonometry where we have to all the value of $x \in [0,2\pi ]$
So we first solve modulus function so if $|x| = a$ then it become $x = \pm a$ and we use some trigonometric relation
$1.{\cos ^2}x - {\sin ^2}x = \cos 2x$
$2.{\sin ^2}x + {\cos ^2}x = 1$ and we rearrange the whole equation by squaring both sides and after that we use these formulas to find our answer.

Step 1. Solve modulus function first
$\sqrt {2{{\sin }^4}x + 18{{\cos }^2}x} - \sqrt {2{{\cos }^4}x + 18{{\sin }^2}x} = \pm 1$
Now doing rearrangements we get
$\sqrt {2{{\sin }^4}x + 18{{\cos }^2}x} = \pm 1 + \sqrt {2{{\cos }^4}x + 18{{\sin }^2}x}$
Now by taking square both side we get
${(\sqrt {2{{\sin }^4}x + 18{{\cos }^2}x} )^2} = {( \pm 1 + \sqrt {2{{\cos }^4}x + 18{{\sin }^2}x} )^2}$
Now ${(a \pm b)^2} = {a^2} + {b^2} \pm 2ab$
By using this we can write
$2{\sin ^4}x + 18{\cos ^2}x = 1 + 2{\cos ^4}x + 18{\sin ^2}x \pm 2\sqrt {2{{\cos }^4}x + 18{{\sin }^2}x}$
Now rearranging this equation
$2{\sin ^4}x - 2{\cos ^4}x + 18{\cos ^2}x - 18{\sin ^2}x = 1 \pm 2\sqrt {2{{\cos }^4}x + 18{{\sin }^2}x}$
Now use some formula
${a^2} - {b^2} = (a - b)(a + b)$ and ${\cos ^2}x - {\sin ^2}x = \cos 2x$
From this we get
$2({\sin ^2}x - {\cos ^2}x)({\sin ^2}x + {\cos ^2}x) + 18({\cos ^2}x - {\sin ^2}x) = 1 \pm 2\sqrt {2{{\cos }^4}x + 18{{\sin }^2}x}$
As we know ${\sin ^2}x + {\cos ^2}x = 1$
Now $- 2\cos 2x + 18\cos 2x = 1 \pm 2\sqrt {2{{\cos }^4}x + 18{{\sin }^2}x}$
From this we can write
$16\cos 2x = 1 \pm 2\sqrt {2{{\cos }^4}x + 18{{\sin }^2}x}$
Now we use $2{\cos ^2}x = 1 + \cos 2x$ and $2{\sin ^2}x = 1 - \cos 2x$
$16\cos 2x = 1 \pm 2\sqrt {\dfrac{1}{2}{{(\cos 2x + 1)}^2} + 9(1 - \cos 2x)}$
Now again we do rearrangements of terms
$16\cos 2x - 1 = \pm 2\sqrt {\dfrac{1}{2}{{(\cos 2x + 1)}^2} + 9(1 - \cos 2x)}$
Now take square both side
${(16\cos 2x - 1)^2} = 4\{ \dfrac{1}{2}({\cos ^2}2x + 1 + 2\cos 2x) + 9 - 9\cos 2x\}$
Now open square and multiply 4 inside the curly braces
$256{\cos ^2}2x + 1 - 32\cos 2x = 2{\cos ^2}2x + 2 + 4\cos 2x + 36 - 36\cos 2x$
Now after rearranging we get
$254{\cos ^2}2x = 37$
We can write this as
$\cos 2x = \pm \sqrt {\dfrac{{37}}{{254}}}$
Now
$2x = {\cos ^{ - 1}}( \pm \sqrt {\dfrac{{37}}{{254}}} )$
$x = \dfrac{1}{2} \times {\cos ^{ - 1}}( \pm \sqrt {\dfrac{{37}}{{254}}} )$ And
Now as question said $x \in [0,2\pi ]$
In this interval $\cos x$ take two time negative value and two time positive value
When $x \in \left[ {0,\dfrac{\pi }{2}} \right]$ , $\cos x$ take positive value so here $x = \dfrac{1}{2} \times {\cos ^{ - 1}}\left( {\sqrt {\dfrac{{37}}{{254}}} } \right)$
When $x \in \left[ {\dfrac{\pi }{2},\pi } \right]$, $\cos x$ take negative value so here $x = \dfrac{1}{2} \times {\cos ^{ - 1}}\left( { - \sqrt {\dfrac{{37}}{{254}}} } \right)$
And also when $x \in \left[ {\pi ,\dfrac{{2\pi }}{3}} \right]$, $\cos x$ take positive value so here $x = \dfrac{1}{2} \times {\cos ^{ - 1}}\left( {\sqrt {\dfrac{{37}}{{254}}} } \right)$
And also when $x \in \left[ {\dfrac{{2\pi }}{3},2\pi } \right]$, $\cos x$ take negative value so here $x = \dfrac{1}{2} \times {\cos ^{ - 1}}\left( { - \sqrt {\dfrac{{37}}{{254}}} } \right)$
From this we can say total 4 solution we have in $x \in \left[ {0,2\pi } \right]$
Note: We have to remember that $\cos \theta$ taking positive value in first and fourth coordinate and negative value in second and third coordinate.