Answer "and" Explanation:
You know that [tex]sin^{2} + cos^{2} x = 1[/tex], so with the given information, you can write
[tex](\frac{7}{10} cos/x-r)^{2} + cos^{2} x= 1[/tex]
that becomes
[tex]\frac{149}{100} cos^{2} / x- \frac{7}{5} r / cos /x + r^{2} - 1 = 0[/tex]
or as well
[tex]149cos^{2} / x -140r / cos / x + 100(r^{2} - 1) = 0[/tex]
The condition for this equation to have real roots is
[tex]70^{2} r^{2} - 149 x 100^{2} (r^{2} - 1) \geq 0[/tex]
hence [tex]r^{2} \leq 149/ 100[/tex]
The roots of the quadratic equation [tex]149t^{2} - 140rt + 100(r^{2} - 1) = 0[/tex] are in the interval [- 1, 1], because with the limitation [tex]|r| \leq \sqrt{149/100}[/tex], the point of a minimum of the polynomial lies between - 1 and 1. Moreover, the polynomial evaluated at - 1 and 1 is > 0 for every r.
Solve for cos x and find the value of sin x.
