Find all horizontal tangents to y=sin(3x)?

Horizontal tangents simply means the slope of line tangent to the function at at that point is equal to 0. From calculus, we know to find the slope of the line tangent to a function at a point, we need to find the derivative at that point. In this problem, we are figuring out when y' is equal to 0, i.e. at what values of x.

y = sin(3x)

y' = 3cos(3x)

3cos(3x) = 0

cos(3x) = 0

From trigonometry, cos(kx) is equal to zero, when: kx = (2npi + pi)/2, solving for x: x = (2npi + pi)/2k, n = 0, +/-1, +/-2,...

3x = (2npi + pi)/2

x = (2npi + pi)/6, n = 0,+/-1, +/-2,...

Differentiate with know to x: 3x^2 + 3y^2dy/dx = 3xdy/dx + 3y sparkling up for dy/dx: dy/dx = (3y - 3x^2)/(3y^2 - 3x) = 0 whilst y = x^2. Now exchange y = x^2 into the unique equation and sparkling up for x: x^3 + x^6 = 3x^3 x^6 - 2x^2 = x^3(x^3-2) = 0. the only answer of this equation it incredibly is smart is x = 2^a million/3, on the grounds that dy/dx is undefined at x = 0, y = 0. for this reason a horizontal tangent happens at x = 2^(a million/3), y = 2^(2/3), and the equation of the horizontal tangent is y = 2^(2/3)

y= +1

&

y = -1

Just look at the sin function:

its a "wavy line" between plus1 and minus 1

d sin(fx) = -f'(x) cos f(x)

SIN X = +/- 1 when - COS X=0