0
Add your answer:
Earn +20 pts
How can I solve tan of 3 theta -sqrt of 3 equals 0 for theta?
tan3A-sqrt3=0 tan3A=sqrt3 3A=tan^-1(sqrt3) 3A= pi/3+npi A=pi/9+npi/3 n=any integer
How do you solve x2 81 equals 0?
x: x2 - 81 = 0
How do you solve 27.82 equals X squared divided by Sin squared x?
Here is a way to solve 27.82 = X2/SIN2(X) using successive approximations or bracketing: First, take the square root of each side: 5.27 = X/SIN(X) SIN(X) has values from 0 to 1 in the first quadrant (0 to 90 deg.) and from 1 to 0 in the second quadrant (90 to 180 deg.) as seen in a table of trig functions. To bracket the answer, plug in values for X in radians; 90 deg. = Pi/2 radians where Pi =3.1415. Pi/2 radians = 1.571 For values of X from 0 to Pi/2, the value of X is too small, so X > 1.571. Since SIN(X) is negative for angles of 180 to 360 deg., the answer should lie between X = 1.571 and X = 3.1415 (Pi radians or 180 deg.) Next, try an answer that lies halfway between 1.571 and 3.1415: Let X = (1.571+3.1415)/2 = 2.356 and solve for X/SIN(X) = 3.33 Since this answer is too small, X must lie between 2.356 and 3.1415 If we continue to bracket the value of X, after a few more steps, we find that if X = 2.621, X/SIN(X) = 5.269 which rounds up to 5.27 This problem can be more easily solved by setting it up in an Excel spread sheet and simply plugging in values for X to converge on the answer.
What is the lim of h if it equals 0 Sinxcosh plus cosxsinh minus sinx divided by h?
lim(h→0) (sin x cos h + cos x sin h - sin x)/h As h tends to 0, both the numerator and the denominator have limit zero. Thus, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply l'Hopital's Rule and the limit equals: lim(h→0) (sin x cos h + cos x sin h - sin x)/h = lim(h→0) (sin x cos h + cos x sin h - sin x)'/h' = lim(h→0) [[(cos x)(cos h) + (sin x)(-sin h)] + [(-sin x)(sin h) + (cos x)(cos h)] - cos x]]/0 = cosx/0 = ∞
Solve the quadratic equation y2-15 equals 0?
y=±√15
Y equals 4 sin x for x equals pi?
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
Can you solve sin 2 theta equals 0?
If sin2(theta) = 0, then theta is N pi, N being any integer
Sin theta equals zero answer in radians in terms of pi?
Theta equals 0 or pi.
How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
What is sin of -pi?
sin(-pi) = sin(-180) = 0 So the answer is 0
What is the sin of pi?
sin(pi) = 0
What is the solution to cos tan-sin over cot equals 0?
either cos OR tan-sin equals zero socos=0 at pi/2 and 3pi/2ortan=sin which is impossibleim not sure though
Find the values of θ in the range 0 to 2π for cos θ plus cos 3θ equals sin θ plus sin 3θ?
The angle can be 0, pi/2, pi, 3*pi/2 or 2*pi radians.
What is the derivative of cos pi x plus sin pi y all to the 8th power equals 44?
(cos(pi x) + sin(pi y) )^8 = 44 differentiate both sides with respect to x 8 ( cos(pi x) + sin (pi y ) )^7 d/dx ( cos(pi x) + sin (pi y) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (-sin (pi x) pi + cos (pi y) pi dy/dx ) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (pi cos(pi y) dy/dx - pi sin (pi x) ) = 0 cos(pi y) dy/dx - pi sin(pi x) = 0 cos(pi y) dy/dx = sin(pi x) dy/dx = sin (pi x) / cos(pi y)
Sin2x equals cos3x find x?
0. sin 2x = cos 3x 1. sin 2x = sin (pi/2 - 3x) [because cos u = sin (pi/2 - u)] 2. [...]
If x is obtuse and sin x equals b how do you show sin 2x is less than b?
you need to explain it better but with what i got i if sin x equals b obviously sin 2x is double of b hence sin 2x is more than b. {Not obvious at all, actually. And the above is false. Sin(Pi/2) = 1 and Sin(Pi)=0. But clearly 2 is not greater than 0. Contradiction.} Obtuse means Pi/2 < x < Pi So, sin(x) = b means b>0, because sin(y) > 0 if 0<y<Pi sin(2x)=2sin(x)cos(x) cos(x) < 0 because cos(Pi/2)= 0 and the derivative is negative there. Hence, sin(2x) = 2 sin(x) cos(x) = 2*b*(-K), where K is a positive constant Since b>0, -2Kb < b
What is the sin of 2 pie?
sin(2*pi) - not pie - is the same as sin(0) = 0
