How do you solve for #A# in this question: #0 <= A <= 360: 2(cosA)^2 + 5sinA - 4 =0#?

1 Answer
Apr 12, 2018

#A=30^0,150^0#

Explanation:

Here,

#2color(blue)(cos^2A)+5sinA-4=0#

#=>2color(blue)((1-sin^2A))+5sinA-4=0#

#=>2-2sin^2A+5sinA-4=0#

#=>2sin^2A-5sinA+2=0#

Factor #color(red)( to[(-4)+(-1)=-5and (-4)xx(-1)=4]#

#=>2sin^2Acolor(red)(-4sinA-sinA)+2=0#

#=>2sinA(sinA-2)-1(sinA-2)=0#

#=>(2sinA-1)(sinA-2)=0#

#=>2sinA-1=0 or sinA-2=0#

#=>2sinA=1 orsinA=2!in[-1,1]=>sinA!=2#

#=>sinA=1/2>0...toI^(st)and II^(nd) Quadrant.#

Given that, #0 <= A <= 360^0#.

#=>A=30^0 ,150^0#