How do you solve $cos (t) = - \frac{\sqrt{3}}{2}$ $cos(t)= -sqrt(3)/2$ over the interval 0 to 2pi?

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How do you solve $cos (t) = - \frac{\sqrt{3}}{2}$ $cos(t)= -sqrt(3)/2$ over the interval 0 to 2pi?

1 Answer

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Answer 1 · 2016-02-20T06:40:01

$t = \frac{5 π}{6} = 150^{\circ}$ $t=(5pi)/6=150^@$ and also $t = \frac{7 π}{6} = 210^{\circ}$ $t=(7pi)/6=210^@$

Explanation:

in the interval $0$ $0$ to $2 π$ $2pi$ , cos t is negative at the 2nd and 3rd quadrants

the reference angle is the special angle $30^{\circ}$ $30^@$

the angle $150^{\circ}$ $150^@$ has a cosine of $- \frac{\sqrt{3}}{2}$ $-sqrt3/2$ also
the angle $210^{\circ}$ $210^@$ has a cosing of $- \frac{\sqrt{3}}{2}$ $-sqrt3/2$

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How do you solve $cos (t) = - \frac{\sqrt{3}}{2}$ $cos(t)= -sqrt(3)/2$ over the interval 0 to 2pi?

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How do you solve $cos (t) = - \frac{\sqrt{3}}{2}$ $cos(t)= -sqrt(3)/2$ over the interval 0 to 2pi?