solving trigonometric equation $3^{\sin^2(x)}+3^{\cos^2(x)}=4$

Question

Please help me to solve this trigonometric equation.

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

Answer 1

As $\sin^2x+\cos^2x=1$

Let $3^{\cos^2x}=a$

$\displaystyle 3^{\sin^2x}=3^{1-\cos^2x}=\frac3{3^{\cos^2x}}=\frac3a$

So,we have $\displaystyle \frac3a+a=4\iff a^2-4a+3=0\iff a=1,3$

If $a=1,$ $3^{\cos^2x}=1=3^0\iff \cos^2x=0\iff \cos x=0,x=(2n+1)\frac\pi2$

If $a=3,$ $3^{\cos^2x}=3=3^1\iff \cos^2x=1\iff \sin x=0,x=n\pi=2n\frac\pi2$ where $n$ is any integer

So, the answer reduces to any [integral] multiple of $\frac\pi2,$ right?

Answer 2

Hint

Let $t=3^{\sin^2 x}$ then the equation is equivalent to (using $\sin^2 x+\cos^2 x=1$) $$t+\frac{3}{t}=4\iff t^2-4t+3=0, \ t\ne0$$ and the roots are $t_1=1$ and $t_2=3$. Can you take it from here?

Answer 3

$$3^{\sin^2x}+3^{\cos^2x}=4\implies 3^{1-\cos^2x}+3^{\cos^2x}=4\implies$$

Putting $\;a:=\cos^2x$ :

$$\frac3{3^a}+3^a=4\implies 3^{2a}-4\cdot3^a+3=0\iff (3^a-3)(3^a-1)=0\implies$$

$$\implies \begin{cases}3^a=3\implies \color{red}{a=1}&\;or\\{}\\3^a=1\implies \color{red}{a=0}\end{cases}$$

and from here

$$\begin{cases}\cos^2x=1\iff \color{green}{x=n\pi}\;,\;\;n\in\Bbb Z&,\;\;or\\{}\\\cos^2x=0\iff \color{green}{x=\frac{2n-1}2\pi}\;,\;\;n\in\Bbb Z\end{cases}$$

Answer 4

We have: $$3^{\sin^2x}+3^{\cos^2x}=4=3+1=3^1+3^0$$

$$\therefore 3^{\sin^2x}+3^{\cos^2x}=3^1+3^0$$

Comparing LHS and RHS, $$\sin^2 x=1$$ $$\cos^2 x =0$$

Or $$\sin^2 x=0$$ $$\cos^2 x=1$$

$\because \sin 0=\cos 90=0$ and $\cos 0=\sin 90=1$,

We get $x=0$ or $x=90$.

Answer 5

Put $3^{\cos^2x}=a,$ $3^{\sin^2x}=b$. Then $a b=3.$ We got the simple system of equations $$ \left \{ \begin{array}{l} a+b=4,\\ a b=3 \end{array} \right. $$ By solving it we get $a=1,3, b=3,1$. Then you may continue as suggested above.