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Please help me to solve this trigonometric equation.

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

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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?

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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?

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  • $\begingroup$ yes, thank you. $\endgroup$ – user97619 Sep 29 '13 at 14:12
  • $\begingroup$ You're welcome:) $\endgroup$ – user63181 Sep 29 '13 at 14:13
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$$3^{\sin^2x}+3^{\cos^2x}=4\implies 3^{1-\cos^2x}+3^{\cos^2x}=4\implies$$

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

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

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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.

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