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Answers
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 ) =
.
.✌good mrning
Answer:
sorry I don't know
sorry I don't know
sorry I don't know