How do you solve $sin 2 θ - cos θ = 0$ $sin 2theta - cos theta = 0$ between 0 and 2pi?

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How do you solve $sin 2 θ - cos θ = 0$ $sin 2theta - cos theta = 0$ between 0 and 2pi?

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Answer 1 · 2016-02-15T08:43:05

in Degrees: $θ = 30^{\circ}, 90^{\circ}, 150^{\circ}, 270^{\circ}$ $" " " "theta=30^@, 90^@, 150^@, 270^@$
in Radians: $θ = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}$ $" " " "theta=pi/6, pi/2, (5pi)/6, (3pi)/2$

Explanation:

The given: Solve
$sin 2 θ - cos θ = 0$ $sin 2theta-cos theta=0$

because
$sin 2 θ = 2 \cdot sin θ cos θ$ $sin 2theta=2*sin theta cos theta$
it follows

$2 sin θ cos θ - cos θ = 0$ $2 sin theta cos theta-cos theta=0$
factor out $cos θ$ $cos theta$

$cos θ (2 sin θ - 1) = 0$ $cos theta(2 sin theta-1)=0$
Equate both factors to zero

$cos θ = 0$ $cos theta=0$ and $2 sin θ - 1 = 0$ $2 sin theta-1=0$
$θ = {cos}^{- 1} (0) = 90^{\circ}, 270^{\circ}$ $theta=cos^-1 (0)=90^@, 270^@$

also

$2 sin θ = 1$ $2 sin theta = 1$
$sin θ = \frac{1}{2}$ $sin theta = 1/2$
$θ = {sin}^{- 1} (\frac{1}{2}) = 30^{\circ}, 150^{\circ}$ $theta=sin^-1 (1/2)=30^@,150^@$

the solutions are:

$θ = 30^{\circ}, 90^{\circ}, 150^{\circ}, 270^{\circ}$ $theta=30^@, 90^@, 150^@, 270^@$

have a nice day...from the Philippines !

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How do you solve $sin 2 θ - cos θ = 0$ $sin 2theta - cos theta = 0$ between 0 and 2pi?

1 Answer

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