I need the Points of Intersection for $\sec^2(x) = 4\cos^2(x)$. I know that $\sec^2(x)/4$ is equal to $1/4\cos^2(x)$ and that $4\cos^2(x)$ is equal to $4[1 + \cos(2x)]/2$. I am trying to eventually factor out the cosines with a $t=$. That would allow me (I'm thinking) to then use the quadratic formula via "completing the square" and then plug t back into the equation. I'm stuck and any help would be appreciated!
Thanks
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$\begingroup$Actually, just let $u=\cos^2(x)$ so that we have
$$\frac1u=4u$$
which occurs at $u=\pm1/2$. Realizing that $u\ge0$, we simply have $u=1/2$.
Can you take it from here?
Bonus hint: It's a special right triangle.
$\endgroup$ 5 $\begingroup$$$\frac{1}{cos^2(x)}=4cos^2(x)$$ $$1 = 4cos^4(x)$$ $$\frac{1}{4} = (cos^2(x))^2$$ $$±\frac{1}{2}=cos^2(x)$$ Just take the positive value of this, because its not possible to further root the negative root. $$cos(x) = ±\sqrt{\frac{1}{2}}$$ Good day.
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