Solve over the Interval cos(2x)+sin(x)=1 , [0,2pi)

Problem

cos(2*x)+sin(x)=1,[0,2*π)

Solution

1. Apply the double angle identity for cos(2*x) to express the equation in terms of sin(x) using cos(2*x)=1−2*sin2(x)

(1−2*sin2(x))+sin(x)=1

2. Simplify the equation by subtracting 1 from both sides and rearranging the terms.

−2*sin2(x)+sin(x)=0

3. Factor the expression by taking out the common factor sin(x)

sin(x)*(−2*sin(x)+1)=0

4. Set each factor to zero to find the possible values for sin(x)

sin(x)=0

−2*sin(x)+1=0⇒sin(x)=1/2

5. Solve for x within the interval [0,2*π) for each case.

sin(x)=0⇒x=0,π

sin(x)=1/2⇒x=π/6,(5*π)/6

Final Answer

cos(2*x)+sin(x)=1⇒x=0,π/6,(5*π)/6,π

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