Question

Solve csc^2 (x) = 4 in 0 <= x < 2pi. Round answers to two decimal places.

Answer 1

I got pi/6, 5pi/6, 7pi,/6, and 11pi/6. But is it supposed to have only two solutions (in quadrants 1 and 2) since it is positive?

Answer 2

is there a type-0 in the inequality?
I'm thinking you meant 0<=x<2pi

Answer 3

Oops...changed it

Answer 4

k all 4 of your solutions are right

Answer 5

But do all four apply? Because if csc is positive then wouldn't the solutions be in only quadrants 1 and 2?

Answer 6

csc can be positive or negative

Answer 7

csc^2 is positive no matter what

Answer 8

so all 4 solutions apply

Answer 9

right
it doesn't matter if csc is negative/positive, the result of csc^2 is positive
so solutions from quadrants 3 and 4 also apply

Answer 10

for example
\[\csc(\frac{\pi}{6})=2 \implies (\csc(\frac{\pi}{6}))^2=2^2 =4 \\ \\ \csc(\frac{5\pi}{6})=2 \implies (\csc(\frac{5 \pi}{6}))^2=2^2 =4 \\ \csc(\frac{7 \pi}{6})=-2 \implies (\csc(\frac{7 \pi}{6}))^2=(-2)^2=4 \\ \csc(\frac{11\pi}{6})=-2 \implies (\csc(\frac{11 \pi}{6}))^2=(-2)^2=4\]

Answer 11

do you see what I mean ?

Answer 12

all 4 of those when plugged into csc^2(x) give you 4

Answer 13

another method can be this:
\[\Large \csc x = \frac{1}{{\sin x}}\]
so we have to solve this equation:
\[\Large {\left( {\frac{1}{{\sin x}}} \right)^2} - 4 = 0\]

Answer 14

I think I get it now. csc^2 makes all four solutions work.

Answer 15

now, such equation, is equivalent to these ones:
\[\Large \frac{1}{{\sin x}} - 2 = 0,\quad \frac{1}{{\sin x}} + 2 = 0\]

Answer 16

or, after a simplification, to these ones:
\[\Large \begin{gathered}
1 - 2\sin x = 0,\quad 1 + 2\sin x = 0 \hfill \\
\hfill \\
\sin x \ne 0 \hfill \\
\end{gathered} \]

Answer 17

Yes

Answer 18

\[\text{ this equation } \sqrt{\csc(x)}=\sqrt{2} \\ \text{ would result \in } x=\frac{\pi}{6}, \frac{5\pi}{6}\]

Answer 19

Because then we would have csc(x) = 2 if both sides were squared

Answer 20

\[\text{ or even this equation } \csc(x)=2 \text{ would result \in } x=\frac{\pi}{6}, \frac{5\pi}{6}\]

Answer 21

csc^2(x)=4 implies though we have both
csc(x)=2 or csc(x)=-2

Answer 22

csc(x)=2 has solutions in quadrant 1 and 2
while
csc(x)=-2 has solutions in quadrant 3 and 4

Answer 23

So csc(x) = 2 is only positive while csc^2(x) = 4 can be both

Answer 24

I'm not liking that question
while solving in the interval [0,2pi)

we only need need to look at when sin(x) is positive to solve csc(x)=2 which is in quadrants 1 and 2.

we only need to look at when sin(x) is negative to solve csc(x)=-2 which is in quadrants 3 and 4.

to solve csc^2(x)=4 you need to look at both previous sentences because
csc^2(x)=4 implies csc(x)=2 or csc(x)=-2

Answer 25

\[\csc^2(x)=4 \\ \sqrt{\csc^2(x)}=\sqrt{4} \\ |\csc(x)|=2 \\ \csc(x)=2 \text{ or } \csc(x)=-2 \\ \sin(x)=\frac{1}{2} \text{ or } \sin(x)=-\frac{1}{2}\]

Answer 26

\[\sqrt{x^2}=|x|=x \text{ or } -x \text{ depending on the value of } x \\ \sqrt{x^2}=x \text{ if } x \ge 0 \\ \sqrt{x^2}=-x \text{ if } x<0 \\ \text{ an example remember when solving } x^2=9 \\ \text{ we said } x=\pm 3 \text{ since } (3)^2=9 \text{ and } (-3)^2=9 \]

Answer 27

\[\csc^2(x) \text{ is the same as } (\csc(x))^2 \]

Answer 28

\[(\csc(x))^2=4 \\ \text{ replace } \csc(x) \text{ with } u \text{ for a second } \\ \text{ how would you solve } u^2=4 \text{ for } u\]

Answer 29

That would be u = 2 or u = -2

Answer 30

right

Answer 31

replace u again with csc(x)
so you have csc(x)=2 or csc(x)=-2