yes y=sinx is x=arcsiny
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2.5
I presume that sin-1x is being used to represent the inverse sin function (I prefer arcsin x to avoid possible confusion). Make use of the trignometirc relationships: cos2θ + sin2θ = 1 ⇒ cosθ = √(1 - sin2θ) cotθ = cosθ/sinθ = √(1 - sin2θ)/sinθ sin(arcsin x) = x Then: cot(arcsin(x)) = √(1 - sin2(arcsin(x))/sin(arcsin(x)) = √(1 - x2)/x ⇒ cot(arcsin(2/3)) = √(1 - (2/3)2)/(2/3) = √(9/32 - 4/32) ÷ 2/3 = √(9 - 4) x 1/3 x 3/2 = 1/2 x √5
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
There are infintely many posibilities: Any expression that contains (y-4) or a multiple of it in the denominator. Also, the following trigonometric functions: An expression that contains arcsin(x/y) where abs(x) < 4 Similarly arccos Corresponding conditions for arccosec and arcsec.
2 sin(x)2 - sin(x) - 1 = 0 Let Y=sin(x) then the equation is 2*Y2 - Y - 1 =0 Delta = (-1 * -1) - 4 * 2 * -1 = 9 Y = (1 + sqrt(9)) / 4 or Y = (1 - sqrt(9)) / 4 Y = 1 or Y = -1/2 Then x = Arcsin(Y) and (in radians) x = Arcsin(1) = Pi/2 +2*k*Pi or x=Arcsin(-1/2) = -Pi/6 + 2*k*Pi where k is an integer