trigometry equation

[mathe25]

i have to find x ...please help me with exercise under b (second ), c (third) and d (last one)

 

[soroban]

Hello, mathe25!

\(\displaystyle \text{Solve for }x:\;\;\log_{\sin x}(2) + \log_{\cos x}(2) + \log_{\sin x}(2)\!\cdot\!\log_{\cos x}(2) \:=\:0 \)

Note that: .\(\displaystyle \sin x,\cos x\,>\,0\:\text{ and }\:\sin x, \cos x \,\ne\,1\)


Use the identity: .\(\displaystyle \log_ba \:=\:\dfrac{1}{\log_ab}\)

We have: .\(\displaystyle \dfrac{1}{\log_2(\sin x)} + \dfrac{1}{\log_2(\cos x)} + \dfrac{1}{\log_2(\sin x)\cdot\log_2(\cos x)} \:=\:0\)



Multiply by \(\displaystyle \log_2(\sin x)\cdot\log_2(\cos x):\)

. . \(\displaystyle \log_2(\cos x) + \log_2(\sin x) + 1 \:=\:0\)

. . . . .\(\displaystyle \log_2(\cos x) + \log_2(\sin x) \:=\:\text{-}1\)

. . . . . . . . . . .\(\displaystyle \log_2(\sin x\cos x) \:=\:\text{-}1\)

. . . . . . . . . . . .\(\displaystyle \log_2\left(\frac{1}{2}\sin2x\right) \:=\:\text{-}1\)

. . . . . . . . . . . . . . . . \(\displaystyle \frac{1}{2}\sin2x \:=\:2^{-1} \:=\:\frac{1}{2}\)

. . . . . . . . . . . . . . . . . \(\displaystyle \sin2x \:=\:1\)

. . . . . . . . . . . . . . . . . . . \(\displaystyle 2x \:=\:\frac{\pi}{2} + 2\pi n\;\text{ for }n = 0,1,2,3,\hdots\)

. . . . . . . . . . . . . . . . . . . . \(\displaystyle x \:=\:\frac{\pi}{4} + \pi n \)


Since \(\displaystyle \sin x,\cos x\) must be positive: .\(\displaystyle x \:=\:\frac{\pi}{4} + 2\pi n\)

 

[mathe25]

thank u

can u help with others exercises ? please and thank u for the first

 

[soroban]

Hello, mathe25!

Here's the next one . . .

\(\displaystyle b)\;\log_{\cos x}(\sin x) + \log_{\sin x}(\cos x) \:=\:2\)

Note that: .\(\displaystyle \sin x,\cos x\:>\:0\,\text{ and }\,\sin x,\cos x \:\ne\:1\)


Use that formula again: .\(\displaystyle \dfrac{\log(\sin x)}{\log(\cos x)} + \dfrac{\log(\cos x)}{\log(\sin x))} \;=\;2\)


Multiply by \(\displaystyle \log(\sin x)\log(\cos x): \;\;\big[\log(\sin x)\big]^2 + \big[\log(\cos x)\big]^2 \;=\;2\log(\sin x)\,\log(\cos x) \)

. . \(\displaystyle \big[\log(\sin x)\big]^2 + 2\big[\log(\sin x)\big]\big[\log(\cos x)\big] + \big[\log(\cos x)\big]^2 \;=\;0 \)

. . . . . . . . . . . . . . \(\displaystyle \big[\log(\sin x) - \log(\cos x)\big]^2 \;=\;0 \)

. . . . . . . . . . . . . . \(\displaystyle \log(\sin x) - \log(\cos x) \;=\;0\)

. . . . . . . . . . . . . . . \(\displaystyle \log(\sin x) \;=\;\log(\cos x)\)

. . . . . . . . . . . . . . . . . . .\(\displaystyle \sin x \;=\;\cos x\)

. . . . . . . . . . . . . . . . . . \(\displaystyle \dfrac{\sin x}{\cos x} \;=\;1\)

. . . . . . . . . . . . . . . . . . \(\displaystyle \tan x \;=\;1\)

. . . . . . . . . . . . . . . . . . . . . \(\displaystyle x \;=\;\frac{\pi}{4} + 2\pi n\)

 

[soroban]

Hello, mathe25!

There's something wrong with the third one.

\(\displaystyle c)\;\log_{\sin x\cos x}(\sin x) + \log_{\sin x\cos x}(\cos x) \:=\:\frac{1}{4}\)

\(\displaystyle \text{We have: }\;\log_{\sin x\cos x}(\sin x) + \log_{\sin x\cos x}(\cos x) \;=\;\frac{1}{4}\)

. . . . . . . . . . . . \(\displaystyle \underbrace{\log_{\sin x\cos x}(\sin x\cos x)}_{\text{This is 1}} \;=\;\frac{1}{4}\)

. . . . . . . . . . . . . . . . . . . . \(\displaystyle 1 \;=\;\frac{1}{4}\) . ??

 

[mathe25]

sorry

yeah i know...my mistake...this is the third one \log_{\sin x\cos x}(\sin x) * \log_{\sin x\cos x}(\cos x) \...sorry
where is + must be *


. . . . . . . . . . .

 

[soroban]

Hello, mathe25!

\(\displaystyle c)\;\log_{\sin x\cos x}(\sin x) \cdot \log_{\sin x\cos x}(\cos x) \;=\;\frac{1}{4}\)

We have: .\(\displaystyle \dfrac{\log(\sin x)}{\log(\sin x\cos x)}\cdot\dfrac{\log(\cos x)}{\log(\sin x\cos x)} \;=\;\dfrac{1}{4}\)


Multiply by \(\displaystyle 4\big[\log(\sin x\cos x)\big]^2\)

. . \(\displaystyle 4\log(\sin x)\log(\cos x) \;=\;\big[\log(\sin x\cos x)\big]^2\)

. . \(\displaystyle 4\log(\sin x)\log(\cos x) \;=\;\big[\log(\sin x) + \log(\cos x)\big]^2\)

. . \(\displaystyle 4\log(\sin x)\log\cos x) \;=\;\big[\log(\sin x)\big]^2 + 2\log(\sin x)\log(\cos x) + \big[\log(\cos x)\big]^2\)

. . . . . . . . . . . . . . . . .\(\displaystyle 0 \;=\;\big[\log(\sin x)\big]^2 - 2\log(\sin x)\log(\cos x) + \big[\log(\cos x)\big]^2\)

. . . . . . . . . . . . . . . . .\(\displaystyle 0 \;=\;\big[\log(\sin x) - \log(\cos x)\big]^2\)


We have: .\(\displaystyle \log(\sin x) - \log(\cos x) \;=\;0\)

. . . . . . . . . . . . . . . . . \(\displaystyle \log(\sin x) \;=\;\log(\cos x)\)

. . . . . . . . . . . . . . . . . . . . .\(\displaystyle \sin x \;=\;\cos x\)

. . . . . . . . . . . . . . . . . . . . \(\displaystyle \dfrac{\sin x}{\cos x} \;=\;1\)

. . . . . . . . . . . . . . . . . . . . \(\displaystyle \tan x \;=\;1\)

. . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle x \;=\;\dfrac{\pi}{4} + 2\pi n \)

 

[mathe25]

thank u

thank u soroban...can u help with the last one too ))