2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,
{2*sin(x) - 1}*{sin(x) - 2)} = 0
=> sin(x) = 1/2 or sin(x) = 2
The second solution is rejected since sin(x) cannot exceed 1.
The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.
sin7x-sin6x+sin5x
Sin 15 + cos 105 = -1.9045
The proof that sin2A plus sin2B plus sin2c equals 4sinAsinBsinC lies in the fact that (sin 2A + sin 2B + sin 2C) = 4 sinA.sinB.sinC.
y=-10 sin 5x sin 5x=y/-10 x=asin(y/-10)/5
The statement of the problem is equivalent to sin x = - cos x. This is true for x = 135 degrees and x = -45 degrees, and also for (135 + 180n) degrees, where n is any integer.
sin7x-sin6x+sin5x
Sin 15 + cos 105 = -1.9045
The proof that sin2A plus sin2B plus sin2c equals 4sinAsinBsinC lies in the fact that (sin 2A + sin 2B + sin 2C) = 4 sinA.sinB.sinC.
y=-10 sin 5x sin 5x=y/-10 x=asin(y/-10)/5
The statement of the problem is equivalent to sin x = - cos x. This is true for x = 135 degrees and x = -45 degrees, and also for (135 + 180n) degrees, where n is any integer.
sex plus sin equals to lust
There is nothing to solve in this equation because there is no =. If you accidentally omitted what the expression equals then resubmit it and I'll be happy to look at it
Assuming the angles are expressed in radians:sin(5x) + sin(x) = 0∴ sin(5x) = -sin(x)∴ 5x = x + π∴ x = π/4On the other hand, if your angles are in degrees, then the answer would be:sin(5x) + sin(x) = 0∴ sin(5x) = -sin(x)∴ 5x = x + 180∴ x = 180°/4∴ x = 45°
leonhard euler
2 sin2 x + sin x = 1. Letting s = sin x, we have: 2s2 + s - 1 = (2s - 1)(s + 1) = 0; whence, sin x = ½ or -1, and x = 30° or 150° or 270°. Or, if you prefer, x = π/6 or 5π/6 or 3π/2.
Well, darling, if we square the first equation and the second equation, add them together, and do some algebraic magic, we can indeed show that a squared plus b squared equals 89. It's like a little math puzzle, but trust me, the answer is as sassy as I am.
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,