hii dear friends...
help me out.....
wanted a solution of a sum....
Find the general solution of cotx + tanx = 2cosecx
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Answered by
9
hiii Nikki,
As given cotx + tanx = 2cosecx
=>. cosx / sinx + sinx / cosx = 2/sinx
=>. (cos^2x + sin^2x) / sinx cosx = 2/sinx
=>. (cos^2x + sin^2x) = 2cosx
=>. 2cosx = 1
=>. cosx = 1/2
=>. cosx = cos π/3
hence the required general solution is:- x= 2nπ +- π/3, where n€Z
As given cotx + tanx = 2cosecx
=>. cosx / sinx + sinx / cosx = 2/sinx
=>. (cos^2x + sin^2x) / sinx cosx = 2/sinx
=>. (cos^2x + sin^2x) = 2cosx
=>. 2cosx = 1
=>. cosx = 1/2
=>. cosx = cos π/3
hence the required general solution is:- x= 2nπ +- π/3, where n€Z
Answered by
12
cotx + tanx = 2cosecx
cosx/sinx + sinx/cosx = 2/sinx
cos²x+sin²x/sinxcosx = 2/sinx
1/sinxcosx = 2/sinx
2sinx cosx - sinx = 0
sinx(2cosx-1) = 0
sinx = 0
or 2cosx - 1 = 0
sinx = 0
or cosx = 1/2
x = nπ
or x = 2nπ ± π/3, n∈z
cosx/sinx + sinx/cosx = 2/sinx
cos²x+sin²x/sinxcosx = 2/sinx
1/sinxcosx = 2/sinx
2sinx cosx - sinx = 0
sinx(2cosx-1) = 0
sinx = 0
or 2cosx - 1 = 0
sinx = 0
or cosx = 1/2
x = nπ
or x = 2nπ ± π/3, n∈z
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