Grade 11Trigonometry
The no.of solytions of tan x + sec x= 2 cos xin [0 ,2π]
1 Answer
Arun
Dear Narendra,
tanx+sec x=2cosx => sin x/cos x + 1/cosx=2 cosx =>(sinx+1)/cosx = 2cosx => 1+sinx=2 cos^2 x
=>1+sin x=2(1-sin^2 x) =>1+sinx -2(1+sinx)(1-sinx)=0 => (1+sinx)(1-2(1-sinx))=0 =>(1+sinx)(2sinx-1)=0
so sin x=1, sin x=1/2
so in between [0,2pi], the solutions are x=3pi/2, x=pi/6, x=5pi/6.
therefore there are 3 solutions.
Cheers!!
Regards,
Arun (askIITians forum expert)
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