Question #e4d1e

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Question #e4d1e

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Answer 1 · 2017-03-14T01:41:51

You just need to simplify the equation.
Here are the steps, from the beginning to the second line of your problem.

Explanation:

You have the following equation:
${sin}^{2} θ - 2 sin θ - 1 = 0$ $sin^2 theta - 2 sin theta -1 =0$

Note that ${sin}^{2} θ = {(sin θ)}^{2}$ $sin^2 theta = (sin theta)^2$ ,
so let's say $sin θ = x$ $sin theta = x$ and rewrite the equation.

We get:
$x^{2} - 2 \cdot x - 1 = 0$ $x^2 - 2*x - 1 = 0$

This is a quadratic equation that you should know how to solve, using the (in)famous quadratic formula:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b \pm sqrt(b^2 -4ac))/(2a)$
where $a$ $a$ , $b$ $b$ , and $c$ $c$ are the coefficients of the second degree polynomial:
$a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$

In our case,
we readily see that we have
$a = 1$ $a=1$ ,
$b = - 2$ $b=-2$ , and
$c = - 1$ $c=-1$

So plugging-in these numbers in the quadratic formula gives the following:
$x = \frac{+ 2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot 1 \cdot (- 1)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 4}}{2}$ $x = (+2 \pm sqrt((-2)^2 - 4*1*(-1)))/(2*1) = (2\pmsqrt(4+4))/2$
so
$x = \frac{2 \pm \sqrt{8}}{2}$ $x = (2\pmsqrt(8))/2$
(Note that this is representing the two solutions of the equation with $x_{1} = \frac{2 + \sqrt{8}}{2}$ $x_1 = (2 + sqrt(8))/2$ and $x_{2} = \frac{2 - \sqrt{8}}{2}$ $x_2 = (2 - sqrt(8))/2$ written together with the $\pm$ $\pm$ ("plus or minus" sign) ).

Replacing the x back to its original form:
$sin θ = \frac{2 \pm \sqrt{8}}{2}$ $sin theta = (2\pmsqrt(8))/2$

So here we are at the second line. I hope this helped.

Answer 2 · 2017-03-14T01:57:29

${sin}^{2} θ - 2 sin θ - 1 = 0$ $sin^2theta-2sintheta-1=0$

We know that the roots of general quadratic equation of the form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ are

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-bpmsqrt(b^2-4ac))/(2a)$

Applying this to our quadratic equation of $sin x$ $sinx$ we get

$sin x = \frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot 1 \cdot (- 1)}}{2 \cdot 1}$ $sinx=(2pmsqrt((-2)^2-4*1*(-1)))/(2*1)$

$\Rightarrow sin x = \frac{2 \pm \sqrt{8}}{2}$ $=>sinx=(2pmsqrt8)/2$

$\Rightarrow sin x = \frac{2 \pm \sqrt{2 \cdot 2^{2}}}{2}$ $=>sinx=(2pmsqrt(2*2^2))/2$

$\Rightarrow sin x = \frac{2 \pm 2 \sqrt{2}}{2}$ $=>sinx=(2pm2sqrt2)/2$

$\Rightarrow sin x = 1 \pm \sqrt{2}$ $=>sinx=1pmsqrt2$

As $sin x = 1 + \sqrt{2} > 0 not possible$ $sinx=1+sqrt2>0" not possible"$

we have

$\Rightarrow sin x = 1 - \sqrt{2} = - 0.414 \approx - {sin 24}^{\circ}$ $=>sinx=1-sqrt2=-0.414~~-sin24^@$

So either $sin x = sin (180 + 24) = {sin 204}^{\circ}$ $sinx=sin(180+24)=sin204^@$

$\Rightarrow sin x = 204^{\circ}$ $=>sinx=204^@$

Or

$sin x = sin (360 - 24) = {sin 336}^{\circ}$ $sinx=sin(360-24)=sin336^@$

$\Rightarrow x = 336^{\circ}$ $=>x=336^@$

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