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

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}

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skopie skopie · Answer 2 · 2019-07-15T11:25:17-04:00

Answer:

Step-by-step explanation:

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

\(2sin^2x-3sinx+1=0\\(2sinx-1)(sinx-1)=0\\2sinx-1=0 \\and\\ sinx-1=0 \\sinx=1 \\and\\sinx=\frac{1}{2}\)

so now we can solve each of these

\(sinx=1\\x=\frac{\pi }{2}\)

and

\(sinx=\frac{1}{2} \\x=\frac{\pi }{6} ,\frac{5pi}{6}\)

solution is

\(x=\frac{\pi }{6},\frac{\pi }{2} ,\frac{5pi}{6}\)