Solve for x 2sin(2x)+ square root of 3=0

Problem

2*sin(2*x)+√(,3)=0

Solution

1. Isolate the trigonometric term by subtracting √(,3) from both sides and then dividing by 2

2*sin(2*x)=−√(,3)

sin(2*x)=−√(,3)/2

2. Identify the reference angle for which the sine value is √(,3)/2

sin(θ)=√(,3)/2⇒θ=π/3

3. Determine the angles in the interval [0,2*π) where the sine function is negative, which occurs in Quadrants III and IV.

2*x=π+π/3+2*k*π=(4*π)/3+2*k*π

2*x=2*π−π/3+2*k*π=(5*π)/3+2*k*π

4. Solve for x by dividing the equations by 2 where k is any integer.

x=(2*π)/3+k*π

x=(5*π)/6+k*π

Final Answer

x=(2*π)/3+k*π,(5*π)/6+k*π

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