y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x
y=2 sin(3x) dy/dx = 2 cos(3x) (3) dy/dx = 6 cos(3x)
Sin[x] = Cos[x] + (1/3)
-(4*log(2*cos(4*x)-4*cos(2*x)+3)-3*log(2*cos(4*x)+2)-2*log(2*cos(2*x)+2))/12
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
sec(x) = 2 so cos(x) = 1/2 and so x = pi/3
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x
One solution. (cos x)2 - 2cos x = 3 Factor: (cos x - 3)(cos x + 1)= 0 cos x = {-1, 3} Solve: For cos x = -1, x = 180 deg No solution for cos x = 3
cos x - 0.5 = 0 ⇒ cos x = 0.5 ⇒ x = 2nπ ± π/3
I'm not really sure what you mean by "the solution", but that equation cos = sec - sintan does simplify down to sin^2 + cos^2 = 1 which so happens to be an identity. I'm not sure if that's what you're looking for, but if it is, here are the steps in simplifying it. 1. Convert sec to 1/cos 2. Convert tan into sin/cos and multiply it by sin sintan = sin(sin/cos) = (sin^2)/cos You then have cos = 1/cos - (sin^2/cos) 3. Multiply everything by cos cos^2 = 1 - sin^2 4. And finally, send the sin^2 over to the left side by adding it (since it is being subracted on the right) You should see this sin^2 + cos^2 = 1 which is an identity.
Yes, x = 2 is the solution.
The answer to the math question Cos 5t cos 3t -square root 3 2 - sin 5t cos 3t equals 0. In order to find this answer you will have to find out what each letter is.
Yes. It is THE solution.
The correct equation is a^3 - a^2*cos(a) - 6cos(a) = 0 which gives cos(a) = a^3/(a^2 + 6) This is not a simple equation to solve. In any case, before attempting a solution, it is necessary to know whether a is measured in degrees or radians.
y=2 sin(3x) dy/dx = 2 cos(3x) (3) dy/dx = 6 cos(3x)
Isolate cos (t): cos(t)=1/3. Use a calculator from here because the answer is not an integer or a simple number.