I've tried using $\frac{c}{a} = xy$ and $\frac{-b}{a} = x+y$ but neither seem to be effective in this case. I'm not really sure how else to approach this question.
Find $K$ if $sin(x)$ and $cos(x)$ are two roots of $8x^2+6Kx+2K+1=0$
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trigonometry
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0hint:$$sina + cos a =x_1+x_2=\frac{-b}{a}=\frac {-6k}{8}\\ sina cos a=x_1.x_2=\frac{2k+1}{8}$$ we know $$sin ^2 a+cos ^2 a=1 \\ \to \\(sina + cos a =\frac {-6k}{8})^2\\1+2ina cos a=(\frac{3k}{4})^2\\1+2.\frac{2k+1}{8}=(\frac{3k}{4})^2$$ – 2017-01-14
1 Answers
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Instead of saying $\sin x$, it would be better to use another variable, say $\sin a$. So assuming $\sin a$ and $\cos a$ are the roots, then using Viete relations. \begin{align*} \sin a + \cos a & = -\frac{6K}{8}\\ \sin a \cos a & = \frac{2K+1}{8} \end{align*} From the first equation (upon squaring etc.) we get $$2 \sin a \cos a =\frac{9K^2}{16}-1.$$ Now using the second equation solve for $K$.