Question

How do you solve the following equation `cos x + sin x tan x = 2` in the interval [0, 2pi]?

Answer 1

`x= pi/3, 5pi/3.`

Explanation:

Given eqn. is `cosx+sinxtanx=2` & the soln. is reqd. to be in `[o,2pi].`

Now, `cosx+sinxtanx=2rArr cosx+sinx*sinx/cosx=2 rArr {cos^2x+sin^2x)/cosx=2 rArr 1/cosx =2 rArr cosx=1/2=cos(pi/3)`

Recall that the Eqn. `costheta=cosalpha` has soln. `theta=2kpi+-alpha, k in ZZ.`

In our case, the soln. is, `x=2kpi+-pi/3, k in ZZ.`

Restricting `x in [0,2pi]`, we have, `x= pi/3, 5pi/3.`