Write the number of solutions of the equation `t a nx+s e cx=2cosx` in the interval `[0,2pi]dot`

Explanation

Given:
`tanx+secx=2cosx`
`x in (2n+1)pi/2`
`sinx+1=2cos^2x`
`sinx+1=2(1-sin^2x)`
`sinx+1=2-2sin^2x`
`sinx+1-2+2sin^2x=0`
`sinx-1+2sin^2x=0`
`2sin^2x+sinx-1=0`
`2sin^2x+2sinx-sinx-1=0`
`2sinx(sinx+1)-1(sinx+1)=0`
`(2sinx-1)(sinx+1)=0`
`2sinx-1=0`
`2sinx=1`
`sinx=1/2`
`x=pi/6,(5pi)/6`
`sinx+1=0`
`sinx=-1`
`x=(3pi)/2`
But `x notin (2n+1)(pi)/2`
Hence, number of solutions are two.