Question

Solve the equation: ${{4}^{\sin 2x+2{{\cos }^{2}}x}}+{{4}^{1-\sin 2x+2{{\sin }^{2}}x}}=65$

Answer 1

Hint: Try to convert the 2nd term of LHS similar to the 1st term of LHS.
Use the formula of trigonometry involving sin and cos .
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Then try to substitute some function of x to make the equation look simpler. Then solve it to get that substituted value and then go back to find x.

Complete step-by-step answer:
We are given the equation as:
${{4}^{\sin 2x+2{{\cos }^{2}}x}}+{{4}^{1-\sin 2x+2{{\sin }^{2}}x}}=65$
We know that
$\begin{align}
  & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & {{\sin }^{2}}x=1-{{\cos }^{2}}x\,\,\,\,\,\cdot \cdot \cdot \text{(i)} \\
\end{align}$
Using equation (i) in LHS of our equation we get
$\begin{align}
  & {{4}^{\sin 2x+2{{\cos }^{2}}x}}+{{4}^{1-\sin 2x+2(1-{{\cos }^{2}}x)}}=65 \\
 & \Rightarrow {{4}^{\sin 2x+2{{\cos }^{2}}x}}+{{4}^{3-\sin 2x-2{{\cos }^{2}}x}}=65\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(ii)} \\
\end{align}$
We know that for any real number a > 0 and m, n $\in \mathbb{R}$
${{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(iii)}$
Using the equation (iii) in the equation (ii) we get
${{4}^{\sin 2x+2{{\cos }^{2}}x}}+\dfrac{{{4}^{3}}}{{{4}^{\sin 2x+2{{\cos }^{2}}x}}}=65$
We observe that the first term and denominator of the second term are the same. So substituting this value i.e. ${{4}^{\sin 2x+2{{\cos }^{2}}x}}$with some variable say t gives us
$t+\dfrac{64}{t}=65$
Multiplying it throughout the equation we get a quadratic equation.
$\begin{align}
  & {{t}^{2}}+64=65t \\
 & \Rightarrow {{t}^{2}}-65t+64=0\,\,\,\,\,\cdot \cdot \cdot \text{(iv)} \\
\end{align}$
We see that above equation is quadratic equation which can be easily solved using quadratic formula which says that
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ are the roots of the quadratic equation $a{{x}^{2}}+bx+c=0\,,\,a\ne 0$
Using this quadratic formula to find t from equation (iv)
a = 1 , b = -65 , c = 64
$\begin{align}
  & t=\dfrac{-(-65)\pm \sqrt{{{(-65)}^{2}}-4(1)(64)}}{2(1)} \\
 & \Rightarrow t=\dfrac{65\pm \sqrt{4225-256}}{2} \\
 & \Rightarrow t=\dfrac{65\pm \sqrt{3969}}{2}=\dfrac{65\pm 63}{2} \\
 & \Rightarrow t=\dfrac{65+63}{2}\,\,\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,\,t=\dfrac{65-63}{2} \\
 & \Rightarrow t=64\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,t=1 \\
\end{align}$
Back substituting t with ${{4}^{\sin 2x+2{{\cos }^{2}}x}}$ and solving for both value of t. First taking t = 64
$\begin{align}
  & {{4}^{\sin 2x+2{{\cos }^{2}}x}}=64={{4}^{3}} \\
 & \Rightarrow \sin 2x+2{{\cos }^{2}}x=3\,\,\,\,\,\,\cdot \cdot \cdot \text{(v)} \\
\end{align}$
We know that
$\begin{align}
  & -1\le \sin 2x\le 1\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(vi)} \\
 & -1\le \cos x\le 1 \\
 & \Rightarrow 0\le {{\cos }^{2}}x\le 1 \\
 & \Rightarrow 0\le 2{{\cos }^{2}}x\le 2\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(vii)} \\
\end{align}$
From equation (vi) and (vii) we see that for equation (v) to be true both $\sin 2x$and $2{{\cos }^{2}}x$has to attain their maximum value. That is
$\begin{align}
  & 2{{\cos }^{2}}x=2\,\,\,\,\,\,\,\text{and}\,\,\,\,\,\,\,\,\sin 2x=1 \\
 & \Rightarrow {{\cos }^{2}}x=1 \\
 & \Rightarrow x=2n\pi \,\,\, \\
\end{align}$
But we see that $x=2n\pi $do not satisfy $\sin 2x=1$because $\sin 4n\pi =0$
So it cannot be equal to 64.
Now coming to t = 1
$\begin{align}
  & {{4}^{\sin 2x+2{{\cos }^{2}}x}}=1={{4}^{0}} \\
 & \Rightarrow \sin 2x+2{{\cos }^{2}}x=0 \\
 & \Rightarrow 2\sin x\cos x+2{{\cos }^{2}}x=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\because \sin 2x=2\sin x\cos x] \\
 & \Rightarrow 2\cos x\left( \sin x+\cos x \right)=0 \\
 & \Rightarrow \cos x=0\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\sin x+\cos x=0 \\
 & \Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{2}\,\,\,\,\,\,or\,\,\,\,\,\,\sin x=-\cos x \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \tan x=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\text{Dividing both side by }\cos x] \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow x=n\pi +\dfrac{3\pi }{4} \\
\end{align}$
So,
$x=\left( 2n+1 \right)\dfrac{\pi }{2}\,$ and $x=n\pi +\dfrac{3\pi }{4}$for $n\in \mathbb{Z}$ is the general solution of the given equation.

Note: This is a very tricky question and needs to be solved in this way otherwise the answer may get long and even may not be correct. The end solution requires solving the general trigonometric equation.