Find all solutions in the interval [0, 2pi).
4sin^(2)x - 4sinx + 1 = 0

Question

Find all solutions in the interval [0, 2pi).
4sin^(2)x - 4sinx + 1 = 0

misty1212 · Answer

i think this is a perfect square 
$$(2\sin(x)-1)^2=0$$

misty1212 · Answer

solve 
$$2\sin(x)-1=0$$ 
do you know how to do it?

misty1212 · Answer

if not, let me know and i can show you

anonymous · Answer

so from what you have i got x=pi/6, 5pi/6. is that right?

misty1212 · Answer

$$\sin(x)=\frac{1}{2}$$ 
$$x=\frac{\pi}{6}$$ and oh yes you are right

anonymous · Answer

okay i'm just not sure how you got (2sin(x)−1)^2=0 so can you explain how you got there?

misty1212 · Answer

oh ok

misty1212 · Answer

treat $\sin(x)$ as its own variable 
like say $u$ or you can even use $x$ as long as you switch back, but lets use $u=\sin(x)$ then you have 
$$4u^2-4u+1=0$$

misty1212 · Answer

i just recognized this as a perfect square
but if you do not, you would try to factor it

misty1212 · Answer

eventually (hope it would be quick) you would see this factors as $(2u-1)(2y-1)$ or 
$(2u-1)^2$

misty1212 · Answer

but  i knew it because $$a^2-2ab+b^2=(a-b)^2$$ and i saw it was of that form

misty1212 · Answer

that is probably more than  you wanted to know, but you have a quadratic equation in sine, so you have to solve it like one

anonymous · Answer

no that was really helpful! i've been stuck on this problem for a little while now lol so thank you so much

misty1212 · Answer

you are welcome, glad to help