Question

4 sin^2 x - 4 sin x + 1 = 0 find all solutions in the interval [0,2pi]

Answer 1

`4 \sin^2 x - 4 \sin x + 1 = 0\\ (2\sin x-1)^2=0\\ 2\sin x-1=0\\ 2\sin x=1\\ \sin x=(1)/(2)\\ x=(\pi)/(6) \vee x=(5\pi)/(6)`

Answer 2

Hello.


`\mathsf{\boxed{4 \sin^(2) x - 4 sin x + 1 = 0}}`

In order to ease comprehension, let's replace 'sin x' with 'y'.


`\mathsf{4y^(2) - 4y + 1 = 0}`


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In the interval: [0, 2pi]:

y = pi/6 or 5pi/6

Hope I helped.