Question

Solve on the interval [0,2pi) 2sin^2x-3sinx+1=0

Answer 1

2sin²(x) + 3sin(x) + 1 = 0
2sin(x)(sin(x)+1) + 1(sin(x)+1) = 0
(sin(x)+1)(2sin(x)+1) = 0
sin(x) = -1
x = {3π/2}
2sin(x) + 1 = 0
sin(x) = -1/2
x = {7π/6, 11π/6}
x = {7π/6, 3π/2, 11π/6}

Answer 2

Explanation:

First we need to factor the given equation and then using the zero producr property we ned to equate each factors to 0 .

`2sin^2x-3sinx+1=0\\(2sinx-1)(sinx-1)=0\\2sinx-1=0 \\and\\ sinx-1=0 \\sinx=1 \\and\\sinx=(1)/(2)`

so now we can solve each of these

`sinx=1\\x=(\pi )/(2)`

and

`sinx=(1)/(2) \\x=(\pi )/(6) ,(5pi)/(6)`

solution is

`x=(\pi )/(6),(\pi )/(2) ,(5pi)/(6)`