solvefor x,4cos(x)*sin(y)=1

解答

求解

x, 4 cos (x) \cdot sin (y) = 1

x = arccos (\frac{1}{4 sin ( y )}) + 2 πn, x = - arccos (\frac{1}{4 sin ( y )}) + 2 πn

求解步骤

4 cos (x) sin (y) = 1

两边除以

4 sin (y); y \neq = 2 πn, y \neq = π + 2 πn

4 cos (x) sin (y) = 1

两边除以

4 sin (y); y \neq = 2 πn, y \neq = π + 2 πn

\frac{4 cos ( x ) sin ( y )}{4 sin ( y )} = \frac{1}{4 sin ( y )}; y \neq = 2 πn, y \neq = π + 2 πn

化简

cos (x) = \frac{1}{4 sin ( y )}; y \neq = 2 πn, y \neq = π + 2 πn

cos (x) = \frac{1}{4 sin ( y )}; y \neq = 2 πn, y \neq = π + 2 πn

使用反三角函数性质

cos (x) = \frac{1}{4 sin ( y )}

cos (x) = \frac{1}{4 sin ( y )}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (\frac{1}{4 sin ( y )}) + 2 πn, x = - arccos (\frac{1}{4 sin ( y )}) + 2 πn

x = arccos (\frac{1}{4 sin ( y )}) + 2 πn, x = - arccos (\frac{1}{4 sin ( y )}) + 2 πn

arctan (\frac{1}{4} x) - π - arctan (\frac{x}{100}) = - π

\frac{cot ( x ) + \frac{1}{c o t ( x )}}{cos ( x )} = sec (x)

sec (θ) = 2, (tan (θ) - sin (θ))^{2} + (1 - cos (θ))^{2}

(1 + csc (A)) (1 - csc (A)) = - cot (A)

6 cos^{2} (3 x) - cos (3 x) - 2 = 0