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anonymous
 4 years ago
Solve the following equation, giving the exact solutions which lie in [0, 2π). (Enter your answers as a commaseparated list.)
sin(2x) = sin(x)
anonymous
 4 years ago
Solve the following equation, giving the exact solutions which lie in [0, 2π). (Enter your answers as a commaseparated list.) sin(2x) = sin(x)

This Question is Closed

DeoxNA
 4 years ago
Best ResponseYou've already chosen the best response.0There'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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for some reason... i'm getting x = 0, pi/3/5pi/3 as the answers

DeoxNA
 4 years ago
Best ResponseYou've already chosen the best response.0No, you're right, I'm wrong. Doing it graphically I do get your answers. That's odd...let me think...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but when i plug them in the program says I'm wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mine. because there is a box that says x=

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0NO...you get \[x=0, \frac{ \pi }{ 3 }, \pi, \frac{ 5\pi }{ 6 }, 2\pi\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@DeoxNA erased a solution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0by dividing \[\sin\theta\], you erase values. that's the worst mistake you can do in these questions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sin(2x)\sin(x)=0\] \[2\sin(x)\cos(x)\sin(x)=0\] \[\sin(x)(2\cos(x)1)=0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sin(x)=0\] or \[\cos(x)=\frac{ 1 }{ 2 }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hey Asteck can I ask you another question or DeoxNa

DeoxNA
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah I was thinking about that, like when you get holes in the graph when you simplify rational functions...Well thank you @Azteck .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Given 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0To get tangent... shouldnt i do sine/cos?

DeoxNA
 4 years ago
Best ResponseYou've already chosen the best response.0I'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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im just trying to get Tangent ... I already know what sine and cos are

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Given 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Like I already know Sine and Cosine,.. Idk how to get Tangent

DeoxNA
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It'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

DeoxNA
 4 years ago
Best ResponseYou've already chosen the best response.0Are 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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