0
Assume the expression is:
2sin²(x) + cos²(x) = 1
Use this identity to work out the problem: cos²(x) = 1 - sin²(x)
Then:
2sin²(x) + 1 - sin²(x) = 1
sin²(x) + 1 = 1
sin²(x) = 0
sin(x) = 0
Therefore, the solutions are:
x = πk radians where k belongs to the set of integers
In degrees, we obtain:
x = 180l° where l belongs to the set of integers
Still curious? Ask our experts.
Chat with our AI personalities
Maxine
Rene
LaoAdd your answer:
Earn +20 pts
Find the second derivative of y equals sin sqaured x?
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
Find all degree solutions 2sin2 6x plus 3sin6x plus 1 equals 0?
2sin2(6x) + 3sin(6x) + 1 = 0 Solving the quadratic, sin(6x) = -1 or sin (6x) = -0.5 sin(6x) = -1 => 6x = 45+60n degrees for integer n sin(6x) = -0.5 => 6x = 35+60n or 55+60n degrees for integer n.
Find the solution 3x-3y equals 12 and 3x plus y equals 8?
70000000y=35t
Find solution for N equals 3?
N = 3. That really is all there is to it.
Given 2sin s equals cos s find all possible values of sin s?
2sin s = cos s; whence, 4sin2 s = cos2 s = 1 - sin2s, and 5sin2 s = 1. Therefore sin s = ±√0.2 = 0.4472, approx. Check: cos2 s = 1 - 0.2 = 0.8, whence, cos s = ±√0.8 = ±2√0.2 = 2 sin s.
