Question

secx/2 = cosx/2

Answer 1

What is your question? Simplify? Solve? And if solve then is there a specified interval or a general solution is required?

Answer 2

solve the equation . tnx :)

Answer 3

hint: 1/cos(x/2) = 1/sin(x/2)

Answer 4

So was there an interval given or just a general solution is required? You didn't answer that part of my question.

Answer 5

@zzr0ck3r, that is incorrect! cos(x/2) does not equal 1/sin(x/2)

Answer 6

the interval is [0,2pi)

Answer 7

\[\sec \frac{ x }{ 2 }=\cos \frac{ x }{ 2 }\]
Because the secant and cosine are reciprocal functions,\[\frac{ 1 }{ \cos \frac{ x }{ 2 } }=\cos \frac{ x }{ 2 }\]
Multiply both sides of the equation by cos(x/2) to obtain\[\cos ^{2}\frac{ x }{ 2 }=1\]
\[\cos \frac{ x }{ 2 }=\pm1\]
Now if 0 ≤ x ≤ 2π is the given interval for x, then \[0\le \frac{ x }{ 2 }\le \pi \]is the interval for x/2. Hence we first solve for x/2 in its interval. Thus if \[\cos \frac{ x }{ 2 }=\pm1\]then\[\frac{ x }{ 2 }=0,or \frac{ x }{ 2 }=\pi \]

∴ x = 0 or x = π

Answer 8

OOPS Sorry! The answer is only x = 0. I made a mistake. I just realized that π is not correct.

Answer 9

Do you understand?

Answer 10

yes i guess

Answer 11

woops sorry I read that as csc