Question

find the valor of x in : sec^2 x = 4cos^2 x

Answer 1

the valor of x? :o
Ohh what a brave little x, such valor he has!!

Answer 2

\[\Large\bf\sf \sec^2x\quad=\quad 4\cos^2x\]Value of x? Hmm let's see..

Answer 3

square rootboth sides

1/cos x = sec x

Answer 4

?! that doesnt help me a lot :(

Answer 5

\[\Large\bf\sf \frac{1}{\cos^2x}\quad=\quad 4\cos^2x\]Multiply both sides by cos^2x,\[\Large\bf\sf 1\quad=\quad 4\cos^4x, \qquad\qquad \cos x\ne0\]We have to make this small distinction now that cosine isn't in the denominator anymore.

Answer 6

Then divide by 4,\[\Large\bf\sf \frac{1}{4}\quad=\quad \cos^4x\]

Answer 7

Then take square root a couple times.
You should see one of your special measures from the unit circle after you do that :o

Answer 8

\[1=4\cos ^4x=\left( 2\cos ^2x \right)^2=\left( 1+\cos 2x \right)^2=1+2\cos 2x+\cos ^2 2x\]
\[\cos ^22x+2\cos 2x=0,\cos 2x \left( \cos 2x+2 \right)=0\]
\[\cos 2x=-2,rejected~\because~\left| \cos 2x \right|\le1\]
\[\cos 2x=0=\cos \left( 2n+1 \right)\frac{ \pi }{2 },~where~n~is ~an~integer.\]
\[x=\left( 2n+1 \right)\frac{ \pi }{ 4 }\]

Answer 9

Have we all forgotten how to factor?

sec^2(x) = 4cos^2(x)

1/[cos^2(x)] = 4cos^2(x)

1 = 4cos^4(x)

\(4\cos^{4}(x) - 1 = 0\)

\((2\cos^{2}(x) + 1)(2\cos^{2}(x) - 1) = 0\)

\(\left(2\cos^{2}(x) + 1\right)\left(\sqrt{2}\cos(x) + 1\right)\left(\sqrt{2}\cos(x) - 1\right) = 0\)

Answer 10

ok thanks!!