Question 605981
how do you solve, 4cos^2x-5sinx-5=0
for 0< x < 2&#960;
4(1-sin^2)-5sin-5=0
4-4sin^2-5sin-5=0
-4sin^2-5sin-1=0
4sin^2+5sin+1=0
(4sin+1)(sin+1)=0
4sin+1=0
sinx=-1/4
x=194.48º and 345.52º
or 
sinx+1=0
sinx=-1
x=180º