How do you solve the following equation #6cos^2x = 19sinx + 16# in the interval [0, 2pi]?

1 Answer
Jul 3, 2016

#x=(180^o)+41^o49'=221^o49' in QIII#
#Q IV, x=(360^o)-(41^o49')=318^o11'#,

Explanation:

#6cos^2x=19sinx+16#
#rArr 6(1-sin^2x)=19sinx+16#
#rArr 6sin^2x+19sinx+10=0#
#rArr 6sin^2x+15sinx+4sinx+10=0#
#rArr 3sinx(2sinx+5)+2(2sinx+5)=0#
#rArr (2sinx+5)(3sinx+2)=0#
#rArr sinx =-5/2,# impossible , as, #sinx in [-1,1], sinx=-2/3#
#sinx =-2/3 ~= -0.6667 = -sin(41^o49')#

Since #sinx<0# and, #x in [0,2pi]#, x lies in Quadrant III, IV.

Accordingly, #x=(180^o)+41^o49'=221^o49' in QIII#
For #Q IV, x=(360^o)-(41^o49')=318^o11'#,