How do you solve the trigonometric inequality $-1<=tan(2pi/3-x)<=sqrt(3)$ ?

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$-1<=tan(2pi/3-x)<=sqrt(3)$

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Answer 1 · 2017-02-12T18:22:46

$-1<=tan((2pi)/3-x)<=sqrt3$

$=>tan(-pi/4)<=tan((2pi)/3-x)<=tan(pi/3)$

$=>-pi/4<=(2pi)/3-x<=pi/3$

$=>pi/4>=-(2pi)/3+x>=-pi/3$

$=>(2pi)/3+pi/4>=(2pi)/3-(2pi)/3+x>=(2pi)/3-pi/3$

$=>(11pi)/12>=x>=pi/3$

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