Question #ab74b

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Question #ab74b

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Answer 1 · 2017-07-31T17:45:53

$θ = \frac{π}{2}, \frac{2 π}{3}, \frac{4 π}{3}$ $theta = pi/2, (2pi)/3, (4pi)/3$

Explanation:

$2 sin θ cos θ + sin θ - 2 cos θ - 1 = 0$ $2sin theta cos theta+sin theta - 2cos theta-1=0$

$2 sin θ cos θ - 2 cos θ + sin θ - 1 = 0$ $2sin theta cos theta - 2cos theta +sin theta -1=0$ $\to$ $->$ rearrange the terms

$2 cos θ (sin θ - 1) + 1 (sin θ - 1) = 0$ $2 cos theta (sin theta - 1) +1(sin theta -1)=0$ $\to$ $->$ factor

$(sin θ - 1) (2 cos θ + 1) = 0$ $(sin theta - 1)(2 cos theta +1) = 0$ $\to$ $->$ factor out $sin θ - 1$ $sin theta - 1$

There are two possibilities:

$sin θ - 1 = 0$ $sin theta - 1 = 0$

$sin θ = 1$ $sin theta = 1$
$θ = \frac{π}{2}$ $theta = pi/2$

$2 cos θ + 1 = 0$ $2 cos theta + 1 = 0$

$2 cos θ = - 1$ $2 cos theta = -1$
$cos θ = - \frac{1}{2}$ $cos theta = -1/2$
$θ = \frac{2 π}{3}, \frac{4 π}{3}$ $theta = (2pi)/3, (4pi)/3$

So, $θ = \frac{π}{2}, \frac{2 π}{3}, \frac{4 π}{3}$ $theta = pi/2, (2pi)/3, (4pi)/3$ .

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Question #ab74b

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