Question

Question #e4d1e

Answer 1

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 theta - 2 sin theta -1 =0`

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

We get:
`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=(-b \pm sqrt(b^2 -4ac))/(2a)`
where `a`, `b`, and `c` are the coefficients of the second degree polynomial:
`ax^2 + bx + c = 0`

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

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

Replacing the x back to its original form:
`sin theta = (2\pmsqrt(8))/2`

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

Answer 2

`sin^2theta-2sintheta-1=0`

We know that the roots of general quadratic equation of the form `ax^2+bx+c=0` are

`x=(-bpmsqrt(b^2-4ac))/(2a)`

Applying this to our quadratic equation of `sinx` we get

`sinx=(2pmsqrt((-2)^2-4*1*(-1)))/(2*1)`

`=>sinx=(2pmsqrt8)/2`

`=>sinx=(2pmsqrt(2*2^2))/2`

`=>sinx=(2pm2sqrt2)/2`

`=>sinx=1pmsqrt2`

As `sinx=1+sqrt2>0" not possible"`

we have

`=>sinx=1-sqrt2=-0.414~~-sin24^@`

So either `sinx=sin(180+24)=sin204^@`

`=>sinx=204^@`

Or

`sinx=sin(360-24)=sin336^@`

`=>x=336^@`