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

1 Answer

Jun 22, 2016

#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.#

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