In the first method, the replacement tan(x) = t sqrt(3) gives you the equation: 3 sqrt(3) t^2 + 2 sqrt(3) t - sqrt(3) = 0 Which can be divided by sqrt(3), leaving you with: 3t^2 + 2t - 1 = 0 Which is quite simpler because it has only rational coefficients.
The only trig identities I still remember (been retired for 10+ yrs now) are "cos-sqr + sin-sqr = 1", & "2.sin(x).cos(x) = sin(2x)". All the others, I have to look up. (I've got 'em bookmarked.) I figured there was probably an identity that would make short work of this, but I went straight for the quadratic, & would have gotten it if I hadn't inadvertently swapped the signs of the 0-th & 1st order coefficients. I got the exact opposite of what you did. I got the math right, it's always the arithmetic that F's me up!
You get 4 values for theta. The positive one gives you 2. 1 in the first and other in the 3rd quadrant. The negative root places them in the 2nd and 4th quadrant. 2nd method: you sneaky....😅 I forgot about that identity.
I got one solution like Multiply nr and dr of LHS by 2 2tanx/(1-tan²x)=tan2x tan2x=√3 x=π/6 another solution x=-π/3 is obtained on solving quadratic equation
problem (tan x) / (1-tan² x) = √3/2 tan x = (√3 / 2 ) - (√3 / 2 ) tan² x (√3 / 2 ) tan² x + tan x -(√3 / 2 ) = 0 Let y = tan x (√3 / 2 ) y² + y - (√3 / 2 ) = 0 Use the quadratic formula. y = {-1 ± √[1+4 (√3 / 2 )(√3 / 2 )]} /√3 = {-1 ± √[1+3]} /√3 = {-1 ± 2} /√3 = 1 /√3 , -√3 tan x = 1 /√3 x = π / 6 + N π, N ∈ ℤ tan x = -√3 x = 2 π / 3 + K π, K ∈ ℤ answer x ∈ { π / 6 + N π, 2 π / 3 + K π, ( N, K ∈ ℤ ) }
@fahrenheit2101 No problema! I am 68 years old. I am not a mathematical. I study by myself, I need to practice. In the subway, bus, restaurant, at home when not working, I am always searching for problema. I like more number theory. I apologize if it seems like a criticism.
@@chillwhale07 there are only two roots but then we made it infinite so the work of Solving is done and then it is made infinite solⁿ. It is just a way to solve
Let y = Tan(x) we get y/(1-y^2) = √3 / 2 2 y = √3 (1-y^2) √3 y^2 + 2 y - √3 =0 y = (-2 +/- √(4-4(-√3)(√3)) )/2√3 y = (-2 +/- √16)/2√3 y = (-1 +/- 2)/√3 = {-3/√3 , 1/√3 } Since y > 0 so y=3/√3 is rejected so we get Tan(x) = 1/√3 OR Tan(x) = -3/√3 = -√3 x = tan⁻¹(1/√3) = Pi/6 OR x = tan⁻¹(-√3) = -Pi/3 = -2 Pi/6 x = Pi/6 + N * Pi OR x = = -2 Pi/6 + N * Pi Final answer is : x = (3 N -2) Pi/6 Where N is integer and (N >= 0)
(1) why does y have to be positive? (2) if you only accept the result y= sqrt(3)/3, then you end up with x = pi/6 + N*pi instead of x=pi/6 + N*pi/2 because if you take N = 1 in your answer, y does not equal sqrt(3)/3
In the first method, the replacement tan(x) = t sqrt(3) gives you the equation:
3 sqrt(3) t^2 + 2 sqrt(3) t - sqrt(3) = 0
Which can be divided by sqrt(3), leaving you with:
3t^2 + 2t - 1 = 0
Which is quite simpler because it has only rational coefficients.
If we multiply by 2 we will have the double angle formula for tangent so tan(2x)=sqrt(3) so
2x=Pi/3+k*Pi => x=Pi/6+k*Pi/2 with K an integer
tan 2x = √3
2x = π/3 + nπ
x = π/6 + nπ/2
That’s what I immediately saw when I looked at the equation.
tanx/(1- tan^2(x)) = sqrt(3)/2
tanx/(2 - sec^2(x)) = "
sinx/(2cosx - 1/cosx) = "
sinx/([2cos^2(x) - 1]/cosx) = "
sinx/([1 - 2sin^2(x)]/cosx) = "
sinx*cosx/cos(2x) = "
sin(2x)/2cos(2x) = "
tan(2x)/2 = sqrt(3)/2
tan(2x) = sqrt(3)
tanx = sqrt(3) at π/3 + πk
∴ tan(2x) = sqrt(3) at (π/3 + πk)/2 or π/6 + πk/2
idk why i did it this way but i love the algebra. so cool!
Doubling both sides results in the tangent double angle formula.
The only trig identities I still remember (been retired for 10+ yrs now) are "cos-sqr + sin-sqr = 1", & "2.sin(x).cos(x) = sin(2x)". All the others, I have to look up. (I've got 'em bookmarked.) I figured there was probably an identity that would make short work of this, but I went straight for the quadratic, & would have gotten it if I hadn't inadvertently swapped the signs of the 0-th & 1st order coefficients. I got the exact opposite of what you did. I got the math right, it's always the arithmetic that F's me up!
You get 4 values for theta. The positive one gives you 2. 1 in the first and other in the 3rd quadrant. The negative root places them in the 2nd and 4th quadrant.
2nd method: you sneaky....😅
I forgot about that identity.
I got one solution like
Multiply nr and dr of LHS by 2
2tanx/(1-tan²x)=tan2x
tan2x=√3
x=π/6
another solution x=-π/3 is obtained on solving quadratic equation
1 - tan^2(x) = cos^2(x)/cos^2(x) - sin^2(x)/sin^2(x) = [cos^2(x) - sin^2(x)]/cos^2(x) = cos(2x)/cos^2(x)
Now we can simplify the fraction on the LHS: tan(x)/(1 - tan^2(x)) = [sin(x)/cos(x)]/[cos(2x)/cos^2(x)] = [sin(x)/cos(x)] * [cos^2(x)/cos(2x)] = sin(x) * cos(x) / cos(2x) = 1/2 * [2 * sin(x) * cos(x)]/cos(2x) = 1/2 * sin(2x) / cos(2x) = 1/2 * tan(2x).
So we have 1/2 * tan(2x) = sqrt(3)/2 --> tan(2x) = sqrt(3). Therefore 2x = pi/3 + k*pi, now to divide by 2, we get x = pi/6 + k*pi/2.
A bit simple
Double both sides, rewrite LHS as tan(2x), and invert.
x = pi/6 + n*pi/2
Edit: Seems I used the 2nd method.
*2: tan(2x)=sqrt(3)
2x=60+180*n, n=0,-1,1,-2,2, ...
x=30+90*n
Lets use coffee one time. 😅
4/2=2, not 1.
tan(x)/(1-tan^2(x))=(3)^1/2/(2)
2tan(x)/(1-tan^2(x))=3^1/2
tan(2x)=3^1/2
tan^-1(tan(2x))=tan^-1(3^1/2)
2x=tan^-1(3^1/2)
x=tan^-1(3^1/2)/2
problem
(tan x) / (1-tan² x) = √3/2
tan x = (√3 / 2 ) - (√3 / 2 ) tan² x
(√3 / 2 ) tan² x + tan x -(√3 / 2 ) = 0
Let
y = tan x
(√3 / 2 ) y² + y - (√3 / 2 ) = 0
Use the quadratic formula.
y = {-1 ± √[1+4 (√3 / 2 )(√3 / 2 )]} /√3
= {-1 ± √[1+3]} /√3
= {-1 ± 2} /√3
= 1 /√3 , -√3
tan x = 1 /√3
x = π / 6 + N π, N ∈ ℤ
tan x = -√3
x = 2 π / 3 + K π, K ∈ ℤ
answer
x ∈ { π / 6 + N π,
2 π / 3 + K π,
( N, K ∈ ℤ ) }
I used the first method.
Do not need to solve a quadratic equation. If you multiply both membros for 2 you have;
tan(2x)=sqr(3) só x=30o + k*90o; k E Z.
Did you watch to the end?
@fahrenheit2101 I solve the problem, if i am abre to solve, before watching the solution, to exercice.
@@pedrojose392 That's fair, it just seemed like you were critiquing the video. Apologies for assuming
@fahrenheit2101 No problema! I am 68 years old. I am not a mathematical. I study by myself, I need to practice. In the subway, bus, restaurant, at home when not working, I am always searching for problema. I like more number theory. I apologize if it seems like a criticism.
Tan2x=Tan(60)degree
2x=60
Or x=60/2=30
tanx = -sqrt3
Why do you say "plus-minus" but write "minus-plus"? It doesn't really matter, just asking.
Or tan(2x) = (1/2)😢😢😢😢😢😢😢
welll.. it's just a quadratic equation..
(senxcosx)/(cos²x - sen²x) = (√3)/2
sen(2x)/cos(2x) = √3
sen(2x) = √3cos(2x)
cos²(2x) = 1/4 => cos(2x) = ± 1/2
cos(2x) = 1/2 => sen(2x) = (√3)/2
=> 2x = π/3 + 2kπ => *x = π/6 + kπ*
cos(2x) = -1/2 => sen(2x) = -(√3)/2
=> 2x = 4π/3 + 2kπ => *x = 2π/3 + kπ*
No quadratic equation has infinite solutions
@chillwhale07 Don't be a dumb. It's a trigonometric function embedded in a quadratic equation.
@SidneiMV wdym? it still has infinite solutions
@@chillwhale07 there are only two roots but then we made it infinite so the work of Solving is done and then it is made infinite solⁿ. It is just a way to solve
Let y = Tan(x) we get
y/(1-y^2) = √3 / 2
2 y = √3 (1-y^2)
√3 y^2 + 2 y - √3 =0
y = (-2 +/- √(4-4(-√3)(√3)) )/2√3
y = (-2 +/- √16)/2√3
y = (-1 +/- 2)/√3 = {-3/√3 , 1/√3 } Since y > 0 so y=3/√3 is rejected so we get
Tan(x) = 1/√3 OR Tan(x) = -3/√3 = -√3
x = tan⁻¹(1/√3) = Pi/6 OR x = tan⁻¹(-√3) = -Pi/3 = -2 Pi/6
x = Pi/6 + N * Pi OR x = = -2 Pi/6 + N * Pi
Final answer is :
x = (3 N -2) Pi/6 Where N is integer and (N >= 0)
(1) why does y have to be positive?
(2) if you only accept the result y= sqrt(3)/3, then you end up with x = pi/6 + N*pi instead of x=pi/6 + N*pi/2 because if you take N = 1 in your answer, y does not equal sqrt(3)/3
@@Packerfan130 you are right. I will fix it.
I like your videos, but I could do without the dumb repetitive puns ("tea", "to be", etc.).
Thanks!
I love the puns 😭
I don't like tea
ua-cam.com/video/Ny5y90YVLmE/v-deo.htmlfeature=shared