First, take the inverse sine of both sides of the equation.
That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that!
It can be proven that sin-1(x) = -i*log[i*x + √(1-x2)]
So in this case:
= -i*log[i*6 + √(1-36)] = -i*log[6*i + √(-35)]
= -i*log(11.916*i)
= 1.57 - 2.48*i
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X is 6
It equals -36. To solve replace x with -6, because it is the value given in the problem and then just multiply. 6(-6)= -36.
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
Sin[x] = Cos[x] + (1/3)
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)