How do you solve $8 {sin}^{2} θ + 17 sin θ + 9 = 0$ $8\sin ^ { 2} \theta + 17\sin \theta + 9= 0$ ?

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How do you solve $8 {sin}^{2} θ + 17 sin θ + 9 = 0$ $8\sin ^ { 2} \theta + 17\sin \theta + 9= 0$ ?

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Answer 1 · 2017-04-27T07:22:33

$θ = - 90$ $theta=-90$

Explanation:

Let $sin θ = a$ $sintheta=a$

So we can write

$8 a^{2} + 17 a + 9 = 0$ $8a^2+17a+9=0$

or

$8 a^{2} + 8 a + 9 a + 9 = 0$ $8a^2+8a+9a+9=0$

or

$8 a (a + 1) + 9 (a + 1) = 0$ $8a(a+1)+9(a+1)=0$

or

$(a + 1) (8 a + 9) = 0$ $(a+1)(8a+9)=0$

or

$a + 1 = 0$ $a+1=0$

or

$a = - 1$ $a=-1$

or

$sin θ = - 1$ $sintheta=-1$

or

$θ = - 90$ $theta=-90$ --------Ans1

or

$8 a + 9 = 0$ $8a+9=0$

or

$8 a = - 9$ $8a=-9$

or

$a = - \frac{9}{8}$ $a=-9/8$

or

$sin θ = - \frac{9}{8}$ $sintheta=-9/8$

or

$θ = I n v a l i d$ $theta=Invalid$

So we get $θ = - 90$ $theta=-90$

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How do you solve $8 {sin}^{2} θ + 17 sin θ + 9 = 0$ $8\sin ^ { 2} \theta + 17\sin \theta + 9= 0$ ?

1 Answer

Explanation:

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