#"using the "color(blue)"trigonometric identity"#
#•color(white)(x)cos2x=1-2sin^2x#
#rArr5/2(1-2sin^2x)-1/2=3sinx#
#rArr5/2-5sin^2x-1/2=3sinx#
#"rearrange into standard form for a quadratic"#
#rArr-5sin^2x-3sinx+2=0larrcolor(blue)"standard form"#
#"multiply through by "-1#
#rArr5sin^2x+3sinx-2=0#
#"we now have a quadratic in sin which can be factored"#
#"in the same way as a quadratic in x"#
#"the factors of "5xx-2=-10#
#"which sum to + 3 are + 5 and - 2"#
#color(blue)"split the middle term using these factors"#
#rArr5sin^2x+5sinx-2sinx-2=0#
#rArrcolor(red)(5sinx)(sinx+1)color(red)(-2)(sinx+1)=0#
#"take out a "color(blue)"common factor "(sinx+1)#
#rArr(sinx+1)(color(red)(5sinx-2))=0#
#"equate each factor to zero and solve for x"#
#sinx+1=0rArrsinx=-1rArrx=270^@#
#5sinx-2=0rArrsinx=2/5#
#"since "sinx>0" then x in first/second quadrants"#
#rArrx=sin^-1(2/5)=23.58^@larrcolor(blue)"first quadrant"#
#rArrx=(180-23.58)^@=156.42^@larrcolor(blue)"second quadrant"#
