Question #e4d1e

2 Answers
Mar 14, 2017

Answer:

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.

Mar 14, 2017

#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^@#