Mathematics - Trigonometric Equation Question with Solution | TestHub

Given,

$2\cos \left(\frac{x^2+x}{6}\right)=4^x+4^{-x}$

Now we know that maximum value of $\cos$ function is $1$, so $2\cos \left(\frac{x^2+x}{6}\right)\le 2$ and using $A.M\ge G.M$ we get $4^x+4^{-x}\ge 2$

So, $\mathrm{L}.\mathrm{H}.\mathrm{S}.\le 2. \& \mathrm{R}.\mathrm{H}.\mathrm{s} \ge 2$

Hence $\mathrm{L}.\mathrm{H}.\mathrm{S}.=2 \& \mathrm{R}.\mathrm{H}.\mathrm{S}.=2$

Now equating both to $2$ we get,

$2\cos \left(\frac{x^2+x}{6}\right)=2$ and $4^x+4^{-x}=2$

Now solving $2\cos \left(\frac{x^2+x}{6}\right)=2$

$\Rightarrow \cos \left(\frac{x^2+x}{6}\right)=1$

$\Rightarrow \frac{x^2+x}{6}=0$

$\Rightarrow x^2+x=0$

$\Rightarrow x=0 \text{or} -1 ...........\left(1\right)$

Now solving $4^x+4^{-x}=2$ we get $x=0 .....\left(2\right)$ 

Now from equation $\left(1\right) \& \left(2\right)$ we can see common solution is $x=0$ 

Hence, possible solution is only one.