Question #ec712

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Answer 1 · 2017-01-03T22:45:46

See below.

$2 cos θ = 1$ $2cos theta = 1$ only if $cos θ = \frac{1}{2}$ $cos theta = 1/2$

Going through the special angles, we know that

$cos 0 = 1$ $cos 0 = 1$

$cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ $cos(pi/6) = sqrt3/2$

$cos (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $cos(pi/4) = sqrt2/2$

$cos (\frac{π}{3}) = \frac{1}{2}$ $cos(pi/3) = 1/2$

So there is our first solution.

Continuing once around the circle ( $0 \leq θ < 2 π$ $0 <= theta < 2pi$ ) we get another solution at $x = \frac{5 π}{3}$ $x = (5 pi)/3$ .

So the solutions are

$\frac{π}{3}$ $pi/3$ and $\frac{5 π}{3}$ $(5pi)/3$ .