Solve the equation $cos(pi/6-x)=-1/2$ in the interval $[0,2pi)$ ?

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Answer 1 · 2017-07-10T05:26:58

$(5pi)/6$ and $(3pi)/2$

As $cos((2pi)/3)=-1/2$ , we can write the equation as

$cos(pi/6-x)=cos((2pi)/3)$

and hence genera solution is $pi/6-x=2npi+-(2pi)/3$ , where $n$ is an integer

i.e. $x=2kpi+pi/6+-(2pi)/3$ , where $k$ is an integer (note $k=-n$ ).

and between $[0,2pi)$ , the solution is $pi/6+(2pi)/3=(5pi)/6$

and $2pi+p/6-(2pi)/3=(3pi)/2$

Solve the equation cos(pi/6-x)=-1/2cos(pi/6-x)=-1/2 in the interval [0,2pi)[0,2pi)?