Question

help please
(cosx-sqrt2/2)(secx-1)=0
use reciprocal identity to express the equation involving secant in term of sine,cosine,or tangent.

Answer 1

\[\frac{1}{\cos \theta}=\sec \theta\]

Answer 2

so thats the answer??

Answer 3

No. That's what you use to get the answer.

Answer 4

so 1/cosx=sec?

Answer 5

Just substitute \(\frac{1}{\cos \theta}\) for \(\sec \theta\) in your equation and solve.

Answer 6

sqrt2/sec?

Answer 7

-1/2(sectheta-1)(sqrt2-2cosx=0

Answer 8

@Limitless is that correct?

Answer 9

http://www.wolframalpha.com/input/?i=%28cosx-sqrt2%2F2%29%281%2Fcostheta-1%29%3D0+

Answer 10

is is quite simple its a factorised equation so ou need to solve

\[\cos(x) - \frac{\sqrt{2}}{2}= 0 \]


\[\cos(x) = \frac{\sqrt{2}}{2}\]

\[x = {\pi}{4} ..or... \frac{7\pi}{4} ..... or x = 45 ..or... 315\]
as well as

\[\sec(x) - 1 = 0\]

\[\frac{1}{\cos(x)} = 1 ... .... \cos(x) = 1 .... then .... x = 0, 2\pi ...or x = 0, 360\]

Answer 11

Sorry, I was AFK. One moment.

Answer 12

\[
\begin{align}
\left(\cos (x)-\sqrt{\frac{2}{2}}\right)\left(\sec (x)-1\right)&=\cos(x)\left(\sec (x)-1\right)-\sqrt{\frac{2}{2}}\left(\sec (x)-1\right)\\
&=\cos(x)\sec(x)-\cos(x)-\sqrt{\frac{2}{2}}\sec(x)+\sqrt{\frac{2}{2}}\\
&=\cos(x)\frac{1}{\cos(x)}-\cos(x)-\sqrt{\frac{2}{2}}\frac{1}{\cos(x)}+\sqrt{\frac{2}{2}}\\
&=1-\cos(x)-\sqrt{\frac{2}{2}}\frac{1}{\cos(x)}+\sqrt{\frac{2}{2}}\\
&=-\cos(x)-\frac{\sqrt{\frac{2}{2}}}{\cos(x)}+1+\sqrt{\frac{2}{2}}\\
\text{ or, if you prefer, } &=-\cos(x)-\frac{\sqrt{4}}{2\cos(x)}+\frac{2+\sqrt{4}}{2}
\end{align}
\]