Question

3^2×5^2×2^3/5^3×2^4×3^3
simplify
please solve this step by step ​

Answer 1

Answer:

Given:cot2θ+sinθ3+3=0

\sf\blue{To\:Find\::General\: Solution\:of\:this\: equation}ToFind:GeneralSolutionofthisequation

\sf\red{Solution\::}Solution:

\sf{\cot^{2}\theta\:+\:\frac{3}{\sin\theta}\:+\:3\:=\:0}cot2θ+sinθ3+3=0

We know that cot = cos/sin so put that value over here which will give you the following equation ,

\sf{\frac{\cos^2\theta}{\sin^2\theta}\:+\:\frac{3}{\sin\theta}\:+\:3\:=\:0}sin2θcos2θ+sinθ3+3=0

Now by using the first identity of trigonometry we get here ,

\sf{\cos^2\theta\:+\:\sin^2\theta\:=\:1}cos2θ+sin2θ=1

Or

\sf{\cos^2\theta\:=\:1-\:\sin^2\theta\:}cos2θ=1−sin2θ

Now use this value here ,

\sf{\frac{1\:-\:\sin^2\theta}{\sin^2\theta}\:+\:\frac{3}{\sin\theta}\:+\:3\:=\:0}sin2θ1−sin2θ+sinθ3+3=0

Now multiply LHS with sin²theta

\sf{1\:-\:\sin^2\theta\:+\:3\sin\theta\:+\:3\sin^2\theta\:=\:0}1−sin2θ+3sinθ+3sin2θ=0

\sf{1\:+\:+\:3\sin\theta\:+\:2\sin^2\theta\:=\:0}1++3sinθ+2sin2θ=0

We can write it as ,

\sf{2\sin^2\theta\:+\:2\sin\theta\:+\:\sin\theta\:+\:1\:=\:0}2sin2θ+2sinθ+sinθ+1=0

= (2 sin theta + 1 )( sin theta + 1 ) = 0

= sin theta = - 1/2 or - 1

\sf{\theta\:=\:n\pi\:+\:(-1)^n\:\frac{-\pi}{6}\:and\:\theta\:=\:n\pi\:+\:(-1)^n\:\frac{-\pi}{2}}θ=nπ+(−1)n6−πandθ