What will be the sign of the product, if we multiply 90 negative integers and 9 positive integers?
Answers
Answer:
\sf\pink{Given\::cot^{2}\theta\:+\:\frac{3}{\sin\theta}\:+\:3\:=\:0}Given:cot
2
θ+
sinθ
3
+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}cot
2
θ+
sinθ
3
+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}
sin
2
θ
cos
2
θ
+
sinθ
3
+3=0
Now by using the first identity of trigonometry we get here ,
\sf{\cos^2\theta\:+\:\sin^2\theta\:=\:1}cos
2
θ+sin
2
θ=1
Or
\sf{\cos^2\theta\:=\:1-\:\sin^2\theta\:}cos
2
θ=1−sin
2
θ
Now use this value here ,
\sf{\frac{1\:-\:\sin^2\theta}{\sin^2\theta}\:+\:\frac{3}{\sin\theta}\:+\:3\:=\:0}
sin
2
θ
1−sin
2
θ
+
sinθ
3
+3=0
Now multiply LHS with sin²theta
\sf{1\:-\:\sin^2\theta\:+\:3\sin\theta\:+\:3\sin^2\theta\:=\:0}1−sin
2
θ+3sinθ+3sin
2
θ=0
\sf{1\:+\:+\:3\sin\theta\:+\:2\sin^2\theta\:=\:0}1++3sinθ+2sin
2
θ=0
We can write it as ,
\sf{2\sin^2\theta\:+\:2\sin\theta\:+\:\sin\theta\:+\:1\:=\:0}2sin
2
θ+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)
n
6
−π
andθ=nπ+(−1)
n
2
−π