SOLUTION: Solve the equation   11tan⁡θ=25sin⁡^2θ tan⁡θ for all nonnegative values of θ less than 360∘.

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Question 1068830: Solve the equation
 
11tan⁡θ=25sin⁡^2θ tan⁡θ for all nonnegative values of θ less than 360∘.
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The symbols are not clear.
If it's
11tan = 25sin^2*tan
11tan - 25sin^2*tan = 0
tan*(11 - 25sin^2) = 0
tan = 0
================
theta = 0, 180 degs
------------------------------------
(11 - 25sin^2) = 0
sin^2 = 11/25
sin(theta) = ħsqrt(11)/5
etc

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
theta=0%5Eo and theta=180%5Eo are solutions because they make tan%28theta%29=0
There are four more solutions that make
sin%28theta%29=sqrt%2811%29%2F5 or sin%28theta%29=-sqrt%2811%29%2F5 .
My calculator says that sin%28theta%29=sqrt%2811%29%2F5
corresponds to highlight%28theta=41.55%5Eo%29 .
That means that theta=180%5Eo-41.55%5Eo=highlight%28138.45%5Eo%29 also has sin%28theta%29=sqrt%2811%29%2F5 .
The calculator would also say that theta=-41.55%5Eo has sin%28theta%29=-sqrt%2811%29%2F5 .
That is not a solution, because it is negative,
but we know that -41.55%5Eo is co-terminal with
theta+=+360%5Eo-41.55%5Eo=358.45%5Eo , which is a solution.
And since we know that sin%28360%5Eo-theta%29=-sin%28theta%29 ,
from the solutions that make sin%28theta%29=sqrt%2811%29%2F5 ,
we can get the solutions that make sin%28theta%29=-sqrt%2811%29%2F5 :
theta=360%5Eo-41.55%5Eo=highlight%28358.45%5Eo%29 and
theta=360%5Eo-138.45%5Eo=highlight%28221.55%5Eo%29