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How do you solve the following equation #cosx(2sinx+1)=0# in the interval [0, 2pi]?

1 Answer

Oct 2, 2016

#"The Soln. Set "={pi/2, 7pi/6, 3pi/2, 11pi/6} sub [0,2pi]#.

Explanation:

#cosx(2sinx+1)=0#

#rArr cosx=0, sinx=-1/2#

# cosx=0, and, 0 le x le 2pi rArr x=pi/2, 3pi/2.#

#sinx=-1/2, &, x in [0,2pi] rArr x=pi+pi/6, or, x=2pi-pi/6,#

#"i.e., "x=7pi/6, 11pi/6.#

Thus, the Soln. Set #={pi/2, 7pi/6, 3pi/2, 11pi/6} sub [0,2pi]#.

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