Solve for x sin(x)^2-2sin(x)+1=0

Problem

sin(x)−2*sin(x)+1=0

Solution

1. Identify the equation as a quadratic in terms of sin(x)

2. Factor the quadratic expression u2−2*u+1 where u=sin(x) which is a perfect square trinomial.

(sin(x)−1)2=0

3. Solve for sin(x) by taking the square root of both sides.

sin(x)−1=0

sin(x)=1

4. Determine the values of x that satisfy the equation within one period [0,2*π)

x=π/2

5. Generalize the solution by adding multiples of the period 2*π where n is any integer.

x=π/2+2*n*π

Final Answer

sin(x)−2*sin(x)+1=0⇒x=π/2+2*n*π

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