How do you solve sin^2(x) + sinx

I am not sure how khổng lồ solve this: $$sin^2 (2x) = 2 sin (2x)$$I thought I could rewrite it lượt thích this: $$4 sin^2 (x) cos^2 (x) = 4 sin (x) cos (x)$$and maybe lượt thích this as well: $$ sin (x) cos (x) = 0$$

but I have no idea whether it is correct và how get lớn the solution ..

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thanks for help

HINT:

Let $sin(2x)=t$ & you get a quadratic equation in $t$.

You need lớn calculate roots of, $t^2-2t=0$.

$$sin^2(2x) = 2sin(2x)$$Assign a variable khổng lồ $sin(2x)$ to lớn get a simple quadratic equation.$$t = sin(2x)$$Rewrite the equation.$$t^2 = 2t$$Now, just solve sầu the equation.$$t^2-2t = 0 implies t(t-2) = 0 implies t = 0 ext OR t = 2$$Plug in $sin(2x)$.For $n in igbiglands.combbZ$, we get the following solutions:$$t = 0 implies sin(2x) = 0 implies 2x = (sin^-1 0)+2pi n$$

$$implies egincases2x = 2pi n implies oxedx = pi n\\2x = pi+2pi nimplies oxedx = fracpi2+pi n\endcases$$We can immediately rule out the second option ($t = 2$) because the range of $sin x$ is $y in <-1, 1>$.

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Combining the two solutions, we can reach a single general solution: $$oxedx = fracnpi2$$

HINT

We have

$$sin^2 (2x) = 2 sin (2x)iff sin^2 (2x) - 2 sin (2x)=0 iff sin (2x)cdot =0$$

then reCall that

$$Acdot B=0 iff A=0 ,lor,B=0$$

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