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First, find the inner expression.
Set `cosx=-sqrt3/2 ` because we need to find what value of x yields `-sqrt3/2 `
Use a unit circle to find this value
=> `x=(5pi)/6 `
Draw a triangle,
Given the problem and using cos rules, we know that short leg is 1, the long leg is `-sqrt3 ` and the hypotenuse is 2.
Using the triangle,
=> `2/1 `
=> `2 `
Refer to the link for the triangle.
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