If $sin θ + 4 csc θ + 5 = 0$ $sintheta + 4csctheta + 5 = 0$ , what is the value of $θ$ $theta$ , on $[0, 2 π)$ $[0, 2pi)$ ?

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If $sin θ + 4 csc θ + 5 = 0$ $sintheta + 4csctheta + 5 = 0$ , what is the value of $θ$ $theta$ , on $[0, 2 π)$ $[0, 2pi)$ ?

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Answer 1 · 2016-11-24T13:47:14

By the reciprocal identity $csc β = \frac{1}{sin β}$ $cscbeta = 1/sinbeta$ :

$sin θ + \frac{4}{sin θ} + 5 = 0$ $sintheta + 4/sintheta + 5 = 0$

Put on a common denominator.

$\frac{{sin}^{2} θ + 4 + 5 sin θ}{sin θ} = 0$ $(sin^2theta + 4 + 5sintheta)/sintheta = 0$

${sin}^{2} θ + 5 sin θ + 4 = 0$ $sin^2theta + 5sintheta + 4 = 0$

Let $t = sin θ$ $t = sintheta$ .

$t^{2} + 5 t + 4 = 0$ $t^2 + 5t + 4 = 0$

$(t + 4) (t + 1) = 0$ $(t + 4)(t + 1) = 0$

$t = - 4 and - 1$ $t = -4 and -1$

$sin θ = - 4 and sin θ = - 1$ $sintheta = -4 and sin theta = -1$

However, since the domain of $y = arcsin x$ $y = arcsinx$ are $- 1 \leq x \leq 1$ $-1 ≤ x ≤ 1$ , there is no real solution to $sin θ = - 4$ $sintheta = -4$ .

Hence, the only solution is $270˚$ ( by the unit circle).

Hopefully this helps!

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If $sin θ + 4 csc θ + 5 = 0$ $sintheta + 4csctheta + 5 = 0$ , what is the value of $θ$ $theta$ , on $[0, 2 π)$ $[0, 2pi)$ ?

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If $sin θ + 4 csc θ + 5 = 0$ $sintheta + 4csctheta + 5 = 0$ , what is the value of $θ$ $theta$ , on $[0, 2 π)$ $[0, 2pi)$ ?