Question

Confusing math question

Answer 1

csc(3∝+40)=sec(7∝−70)

Answer 2

Do they mean the limit?

Answer 3

so... what's the question at -> \(\bf csc(3\alpha+40)=sec(7\alpha-70)?\)

Answer 4

Find the ∝

Answer 5

@jdoe0001

Answer 6

3 eqns that can be used to solve the prob.

\[\csc A = \frac{ 1 }{ \sin A }\]
\[\sec A = \frac{ 1 }{ \cos A }\]
and
\[\sin (\frac{ \pi }{ 2 } - A) = \cos A\]

Answer 7

wow

Answer 8

are you supposed to find one answer, or all of them?
one will be no problem, all will be almost impossible

Answer 9

Just one answer, I am confused about tis formula

Answer 10

one answer we can start with
\[\frac{1}{\cos(3x+40)}=\frac{1}{\sin(7x-70)}\]

Answer 11

or
\[\cos(3x+40)=\sin(7x-70)\]

Answer 12

damn i have that backwards!!

Answer 13

\[\sin(3x+40)=\cos(7x-70)\]

Answer 14

x=12, right?

Answer 15

idk i am still working

Answer 16

add \(90\) to \(7x-70\) to get
\[\cos(7x-70)=\sin(7x+20)\]

Answer 17

so one answer will be the solution to
\[3x+40=7x+20\]

Answer 18

i get \(x=5\)