Please help Find the specific values of x on the interval [0,2pi] sin(2x)tan(2x)+sin(2x)=0
I tried converting tan to sin/cos And then simplifying from there by converting everything into sines and multiplying, but I have no idea where to go from here
Umm I need to do this algebraically so I can't just go through an check them I need to simplify the equation
I notice that your expression can be factored. Would you please factor it now.
Oh sorry I was gone Ok so I got sin2x(tan2x+1) Do I expand using addition properties?
\(\color{#000000 }{ \displaystyle ab+a=0 }\) \(\color{#000000 }{ \displaystyle (a\times b)+(a\times 1)=0 }\) \(\color{#000000 }{ \displaystyle a(b+1)=0 }\) \(\color{#000000 }{ \displaystyle a=0 }\) or, \(\color{#000000 }{ \displaystyle b+1=0\quad \Longrightarrow\quad b=-1 }\)\(\small .\)
You are just applying the basic algebra, but to trigonometric functions, rather than just numbers.
Oh I see that Lol I was thinking more complicated Thank you!
Well, what you have at the moment (which is correct) is\(\small : \) \(\color{#0000ff }{ \displaystyle \sin(2x)\left[\tan(2x)+1\right]=0}\) Then, by the "zero property" you have\(\small : \) \(\color{#0000ff }{ \displaystyle \sin(2x)=0\quad}\)or,\(\color{#0000ff }{ \displaystyle \quad \tan(2x)+1=0}\)
Yeah I got the specific solutions of 0, pi/2,pi,3pi/2,2pi,3pi/8,7pi/8,11pi/8, and 15pi/8 Is that correct?
\(\color{#0000ff }{ \displaystyle \sin(2x)=0\quad {\small \rm then}\quad x=0,~~\frac{\pi}{2},~~\frac{3\pi}{2},~~2\pi}\) \(\color{#0000ff }{ \displaystyle \tan(2x)+1=0\quad {\small \rm then}\quad x=\frac{3\pi}{8},~~\frac{7\pi}{8},~~\frac{11\pi}{8},~~\frac{15\pi}{8}}\)
yes, your answers are all correct.
(I am lagging a little, sorry takes time to type it)
Alright thank you!
YW
the first one is for values where sin(x) and cos(x) are zero, what is the second for the double angle of the tangent
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