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Solve the following equation, giving the exact solutions which lie in [0, 2π). (Enter your answers as a comma-separated list.) sin(2x) = sin(x)

Mathematics
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There's a pythagorean identity that says: sin(2x)=2*sin(x)*cos(x). so... 2cos(x)=1 cos(x)=1/2 And here's when your trig flashcards come in handy...
for some reason... i'm getting x = 0, pi/3/5pi/3 as the answers
im i wrong?

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Other answers:

No, you're right, I'm wrong. Doing it graphically I do get your answers. That's odd...let me think...
but when i plug them in the program says I'm wrong
Your's or mine?
mine. because there is a box that says x=
NO...you get \[x=0, \frac{ \pi }{ 3 }, \pi, \frac{ 5\pi }{ 6 }, 2\pi\]
@DeoxNA erased a solution.
by dividing \[\sin\theta\], you erase values. that's the worst mistake you can do in these questions.
oh i see
\[\sin(2x)-\sin(x)=0\] \[2\sin(x)\cos(x)-\sin(x)=0\] \[\sin(x)(2\cos(x)-1)=0\]
\[\sin(x)=0\] or \[\cos(x)=\frac{ 1 }{ 2 }\]
Hey Asteck can I ask you another question or DeoxNa
Yeah I was thinking about that, like when you get holes in the graph when you simplify rational functions...Well thank you @Azteck .
@dainel40 Sure thing...
Given the information below, find the exact values of the remaining circular functions of θ. sec(θ) = 10 with θ in Quadrant IV I got Sine= -sqrt(99)/10 Cos= 1/10 But I can't get Tangent... I tried doing Sine/Cos but I get the wrong answer
To get tangent... shouldnt i do sine/cos?
I'm sorry, but I don't really get what you're doing. Are you trying to convert the secant to the other circular functions? Or do you replace the secant with each respective function?
im just trying to get Tangent ... I already know what sine and cos are
The tangent of what?
Given the information below, find the exact values of the remaining circular functions of θ. sec(θ) = 10 with θ in Quadrant IV I got Sine= -sqrt(99)/10 Cos= 1/10 But I can't get Tangent... I tried doing Sine/Cos but I get the wrong answer
Like I already know Sine and Cosine,.. Idk how to get Tangent
Ok, sorry I didn't get what you were saying at first....Somehow I got it this time, so if: sec(θ)=10 in Quadrant IV θ=4.8122 tan(4.8122)=-9.985. How did you get an exact theta?
It's -√99/10. sin^2 + cos^2 = 1 (1/10)^2 + sin^2 = 1 1/100 + sin^2 = 1 sin^2 = 99/100 sin = ± √ (99/100) = ± √99 / √100 = ± √99 / 10
Are you sure its not (√ 99/10)/(1/10)=√ 99 ? I know I'm not being much help, but its just that the answer makes perfect sense...

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