Question

Find all solutions to the equation csc x (2sinx-sqrt(2)) = 0
A. x= Pi/4 + kPi , where k is any positive integer.
B. x= Pi/4 +2kPI and 7Pi/4 + 2kPI, where k is any positive integer
C. x= 3pi/4 + 2Pik, where k is any positive integer.
D. x = Pi/4 + 2kPi and 3Pi/4+ 2kPi, where k is any positive integer.

Answer 1

@RadEn

Answer 2

\[\csc x (2\sin x-\sqrt2) = 0\]

Answer 3

thats the equation..

Answer 4

\[{(2sinx-\sqrt2) \over \sin x} = 0\]\[{2\sin x \over \sin x}-{\sqrt2 \over \sin x}=0\]\[2-{\sqrt2 \over \sin x}=0\]

Answer 5

\[2={\sqrt2 \over \sin x}\]

Answer 6

\[2 \sin x=\sqrt2\]

Answer 7

im not really seeing what you are saying..

Answer 8

\[\sin x={\sqrt 2 \over 2}\]\[x={\pi \over 4},{3\pi \over 4}\]

Answer 9

csc x (2sinx-sqrt(2)) = 0
for zeroes cscx=0 or 2sinx-sqrt(2)
case I :
cscx=0
1/sinx = 0 ----> no solution for x (it does not exis)
case II :
2sinx-sqrt(2) = 0
2sinx = sqrt(2)
sinx = 1/2 * sqrt(2)
* for sinx = sin(pi/4)
sinx=sin(2kpi+pi/4)
x=2kpi+pi/4
** for sinx=sin(3pi/4)
sinx=sin(2kpi+3pi/4)
x=2kpi+3pi/4

Answer 10

so....

Answer 11

D :p

Answer 12

do you understand why it is D

Answer 13

not really ..id like explaining...i don't really care for just answers..

Answer 14

If you look at my work I prove that the two answers are pi/4 and 3pi/4. Before I explain the k stuff do you understand those steps?

Answer 15

yeah i think so

Answer 16

From the unit circle I knew those solutions. Also one full rotation around the circle is 2pi right?! so pi/4 + 2pi would be right back at the same spot, giving us the same answer. Then they added for any positive integer k because if this continued on for infinity, any positive integer would make just one more rotation

2pi, 4pi, 6pi, and so on. all giving the same value

Answer 17

oh..okay

Answer 18

Hope that helped