Solve for x sin(x)=2

Problem

sin(x)=2

Solution

1. Identify the range of the real-valued sine function. For any real number x the value of sin(x) must fall within the interval [−1,1]

2. Compare the given value to the range. The equation states sin(x)=2 Since 2>1 there are no real solutions for x

3. Apply the definition of the inverse sine function using complex numbers. To find complex solutions, use the exponential form sin(x)=(e(i*x)−e(−i*x))/(2*i)

4. Substitute the value into the exponential equation.

(e(i*x)−e(−i*x))/(2*i)=2

5. Rearrange into a quadratic form. Let u=e(i*x)

u−1/u=4*i

u2−4*i*u−1=0

6. Solve for u using the quadratic formula.

u=(4*i±√(,(4*i)2−4*(1)*(−1)))/2

u=(4*i±√(,−16+4))/2

u=(4*i±√(,−12))/2

u=(4*i±2*i√(,3))/2

u=i*(2±√(,3))

7. Solve for x by taking the natural logarithm. Since e(i*x)=u then i*x=ln(u)+2*k*π*i

x=ln(i*(2±√(,3)))/i+2*k*π

x=−i*ln(i*(2±√(,3)))+2*k*π

8. Simplify using the property ln(a*b)=ln(a)+ln(b) and ln(i)=(i*π)/2

x=−i*(ln(2±√(,3))+(i*π)/2)+2*k*π

x=π/2−i*ln(2±√(,3))+2*k*π

Final Answer

sin(x)=2⇒x=2*k*π+π/2±i*ln(2+√(,3))

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