Question

solve for x:
2 arctan(sinx)=arctan(2secx) x neq to pi/2

Answer 1

\[\Huge 2\tan^{−1}(sinx)=\tan^{-1}(2secx)\]

Answer 2

hmm... The one in your post is different from the one ^ up there.
Are these two different questions?

Answer 3

simplify and equate! :O :O

Answer 4

same now!

Answer 5

\[\LARGE \frac{2sinx}{1+\sin^2x}=\tan^{-1}(\frac{2}{cosx})\]

Answer 6

re-check LHS :3

Answer 7

/??

Answer 8

??

Answer 9

@shubhamsrg

Answer 10

denominator will be 1-sin^2x in LHS

Answer 11

okay,next?

Answer 12

its simply after that
1-sin^2x becomes cos^2x
then it'll get simplified

Answer 13

i am getting tan^2x=1
x=pi/4,-pi/4..so on?

Answer 14

can anyone check

Answer 15

seems right

Answer 16

but ans-pi/4

Answer 17

oh yes,hmm, we'll again put those values back in the original eqn and check which one satisfies

Answer 18

no domain given btw

Answer 19

I am getting pi/4,-pi/4..3pi/4..-3pi/4..like that..
(-1)^n (2n-1)pi/4..

Answer 20

@amistre64 @Callisto @shubhamsrg @experimentX @dan815 @mathslover @terenzreignz

Answer 21

\[\LARGE \frac{2sinx}{cos^{2}x}=\tan^{-1}(\frac{2}{cosx})\]
\[\LARGE \frac{2Tan^{2}}{cosx}=(\frac{2}{cosx})\]
\[\LARGE \ 2Tan^{2}=({2})\]
Tan=1