Solve for x if sin2x=cos(x+30) for x [-180;180]
sin2x = cos(x+30)
when x=120,
sin 240 = cos 150
There are other values, but you have to solve a 4th-degree equation to get them.
2sinx cosx = cos30 cosx - sin30 sinx
square both sides, collect terms, convert sin^2 = 1-cos^2, and square both sides again. Gets messy
To solve the equation sin(2x) = cos(x+30), we can start by using the identities sin(2x) = 2sin(x)cos(x) and cos(x+30) = cos(x)cos(30) - sin(x)sin(30).
The equation becomes 2sin(x)cos(x) = cos(x)cos(30) - sin(x)sin(30).
Next, let's simplify the equation by collecting like terms:
2sin(x)cos(x) - cos(x)cos(30) + sin(x)sin(30) = 0
Now, using the trigonometric identity cos(A+B) = cos(A)cos(B) - sin(A)sin(B), we can simplify the equation further:
2sin(x)cos(x) - cos(x)cos(30) + sin(x)sin(30) = 0
2sin(x)cos(x) - cos(x)*cos(30) + sin(x)*sin(30) = 0
2sin(x)cos(x) - cos(x)*cos(30) + sin(x)*sin(30) = 0
2sin(x)cos(x) - cos(x)*cos(30) + sin(x)*sin(30) = 0
2sin(x)cos(x) - cos(30)*cos(x) + sin(30)*sin(x) = 0
(cos(30) - 2cos(x))cos(x) + sin(30)*sin(x) = 0
Now we have a quadratic equation in terms of cos(x), let's solve for cos(x):
(cos(30) - 2cos(x))cos(x) + sin(30)*sin(x) = 0
(cos(30) - 2cos(x))cos(x) = -sin(30)*sin(x)
(cos(30) - 2cos(x)) = -sin(30)
cos(30) - 2cos(x) = -sin(30)
cos(x) = (cos(30) + sin(30))/2
Now we can find the value of cos(x) using a calculator or by knowing the value of cos(30) and sin(30).
Once we find the value of cos(x), we can find the value of x using the inverse cosine function (cos^(-1)).
Note: Since we are looking for solutions in the range [-180, 180], make sure to get the principal value of the inverse cosine function that falls within this range.