SOLUTION: Please solve the equation: (x is less than or equal to 2pi and greater than or equal to 0) 1.) 3sin^2x-8sinx-3=0 2.) sin^2x+3cos^2x=0

Algebra ->  Trigonometry-basics -> SOLUTION: Please solve the equation: (x is less than or equal to 2pi and greater than or equal to 0) 1.) 3sin^2x-8sinx-3=0 2.) sin^2x+3cos^2x=0       Log On


   

Question 644039: Please solve the equation: (x is less than or equal to 2pi and greater than or equal to 0)
1.) 3sin^2x-8sinx-3=0
2.) sin^2x+3cos^2x=0
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Please solve the equation: (x is less than or equal to 2pi and greater than or equal to 0)
1.) 3sin^2x-8sinx-3=0
(3sinx+1)(sinx-3)=0
..
3sinx+1=0
sinx=-1/3
using sin^-1 key of calculator
x≈0.348 and 5.94 radians in quadrants II and IV where sin<0
or
sinx-3=0
sinx=3>1 (no solution, -1≤sinx≤1)
..
2.) sin^2x+3cos^2x=0
sin^2x+3(1-sin^2x)=0
sin^2x+3-3sin^2x=0
-2sin^2x+3=0
2sin^2x=3
sin^2x=3/2=1.5
sinx=√1.5>1 (no solution, -1≤sinx≤1)