A) 0
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
Given, equation is \[\tan x+\sec x=2\cos x\] \[\Rightarrow \] \[\sin x+1=2{{\cos }^{2}}x\] \[\Rightarrow \] \[2(1-{{\sin }^{2}}x)=\sin x+1\] \[\Rightarrow \] \[2{{\sin }^{2}}x+\sin x-1=0\] \[\Rightarrow \] \[(2\sin x-1)(\sin x+1)=0\] \[\Rightarrow \] \[\sin x=\frac{1}{2}\]or\[\sin x=-1\] \[\Rightarrow \] \[x=\frac{\pi }{6},\frac{5\pi }{6}\] Or \[x=\frac{3\pi }{2}\] \[\Rightarrow \] \[x=\frac{\pi }{6},\frac{5\pi }{6}\] (\[\because \]\[x=\frac{3\pi }{2},\tan x\] is not defined) Hence, in the interval\[[0,2\pi ]\]the number of solutions is 2.
You need to login to perform this action.
You will be redirected in
3 sec
