Gettin' Triggy With The Log

$\begin{aligned} \large \log_{\cos x} \sin x+\log_{\sin x} \cos x&=\large 2 \\ \large \frac{\log {\sin x}}{\log {\cos x}}+\frac{\log {\cos x}}{\log {\sin x}}&=\large 2 \end{aligned}$ Applying the AM-GM inequality , $\large \dfrac{\log {\sin x}}{\log {\cos x}}+\dfrac{\log {\cos x}}{\log {\sin x}}\ge 2\sqrt{\dfrac{\log {\sin x}}{\log {\cos x}}\cdot \dfrac{\log {\cos x}}{\log {\sin x}}}=2$ From the equality condition of AM-GM inequality, we have, $\begin{aligned} \large \dfrac{\log {\sin x}}{\log {\cos x}}&=\large \dfrac{\log {\cos x}}{\log {\sin x}} \\ \large (\log {\sin x})^2&=\large (\log {\cos x})^2\\ \large \sin x &=\large \cos x \end{aligned}$ Therefore, $0<x<\dfrac{\pi}{2}\implies x=\boxed{\dfrac{\pi}{4}}$

Since $\log_ab = (\log_ba)^{-1}$ , we are solving $u + u^{-1} = 2$ , where $u=\log_{\sin x}{\cos x}$ , and hence $\log_{\sin x}\cos x = u = 1$ , and hence $\sin x = \cos x$ , and hence $x = \boxed{\tfrac14\pi}$ for $x \in (0,\tfrac12\pi)$ .

$\log_{\cos(x)}\sin\left(x\right)+\log_{\sin(x)}\cos\left(x\right)=2$ $\frac{\ln\sin\left(x\right)}{\ln\cos\left(x\right)}+\frac{\ln\cos\left(x\right)}{\ln\sin\left(x\right)}=2$ $\frac{\left(\ln\sin\left(x\right)\right)^{2}+\left(\ln\cos\left(x\right)\right)^{2}}{\ln\cos\left(x\right)\ln\sin\left(x\right)}=2$ (neither $\cos(x)$ nor $\sin(x)$ is zero given that the equation's domain excludes 0 and $\frac{\pi}{2}$ ) $\left(\ln\sin\left(x\right)\right)^{2}-2\ln\cos\left(x\right)\ln\sin\left(x\right)+\left(\ln\cos\left(x\right)\right)^{2}=0$ $\left(\ln\sin\left(x\right)-\ln\cos\left(x\right)\right)^{2}=0$ $\ln\sin\left(x\right)=\ln\cos\left(x\right)$ $\sin\left(x\right)=\cos\left(x\right)$ $\boxed{x=\frac{\pi}{4}}$ for the interval $x \in (0, \frac{\pi}{2})$

Log sinx +log cosx=2 cosx sinx I assumed that both of the terms in left side is equal to 1. So that 1+1 =2.

As we know logx(basex)=1. So sinx=cosx It is true for only one value of x in (0,90)

So, x=45 i.e. pi/4.

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We know already two things : 1)Loga(b)=ln(b)/ln(a)

2) sin(pi/4)=cos(pi/4) so for x=pi/4 we have a=b

Therefore We can turn this equation (for x=pi/4) into : Loga(a)+Loga(a)=1+1=2 Therefore x must be equal to pi/4 mod(2pi)

log(cosx) sinx + log(sinx) cosx = 2 log(cosx) sinx + 1/[log(cosx) sinx] = 2 === say log(cosx) sinx = p ==== p + 1/p = 2 ==========×p p²-2p+1=0 (p-1)²=0 p=1 log(cosx) sinx =1 sin x = cos x ¹ x= 45°