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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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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.

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    so the points of intersection are pi /3 and 5pi/3? – 2017-01-18
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    @Billy No. We have $\cos(x)=\pm\sqrt{1/2}=\pm\frac{\sqrt 2}2$... – 2017-01-18
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    @Billy You should've made it to this step:$$\frac1{\cos^2(x)}=4\cos^2(x)$$And it is standard technique to let $u=\text{trig function}$ to simplify, giving$$\frac1u=4u$$Multiply both sides by $u$ and we get quadratic. – 2017-01-18
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    ah yes now I see.Great thank you. But but say I did not realize to use u in the beginning, how would I get the quadratic? I eventually got 1 = 8cos^2(x) * {cos^2(x)[8cos(2x)]}. I do not know how to multiply two cosines when one cosine has an exponent and the other does not. – 2017-01-18
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    @Billy I honestly have no idea what you would do in that scenario. In general, you shouldn't make steps that change the value inside the trig function. That is, no $\cos(2x)$'s around... – 2017-01-18
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$$\frac{1}{cos^2(x)}=4cos^2(x)$$ $$1 = 4cos^4(x)$$ $$\frac{1}{4} = (cos^2(x))^2$$ $$±\frac{1}{2}=cos^2(x)$$ 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.