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

Problem

√(,3)*sec(x)+2=0

Solution

1. Isolate the secant term by subtracting 2 from both sides of the equation.

√(,3)*sec(x)=−2

2. Divide both sides by √(,3) to solve for sec(x)

sec(x)=−2/√(,3)

3. Rewrite the equation in terms of cosine using the reciprocal identity cos(x)=1/sec(x)

cos(x)=−√(,3)/2

4. Identify the reference angle. Since cos(π/6)=√(,3)/2 the reference angle is π/6

5. Determine the quadrants where cosine is negative. Cosine is negative in Quadrant II and Quadrant III.

x=π−π/6=(5*π)/6

x=π+π/6=(7*π)/6

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

x=(5*π)/6+2*n*π

x=(7*π)/6+2*n*π

Final Answer

√(,3)*sec(x)+2=0⇒x=(5*π)/6+2*n*π,(7*π)/6+2*n*π

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