Question

For questions 1-8 verify that the given value is a solution for the trigonometric equation by showing that it makes the eqution true. PLEASE HELP ME OUT I NEED ANSWERS !!

Answer 1

surjithayer @surjithayer

Answer 2

i do not see any question here.

Answer 3

For questions 1-8 verify that the given value is a solution for the trigonometric equation by showing that it makes the eqution true. PLEASE HELP ME OUT I NEED ANSWERS !!

Answer 4

@surjithayer

Answer 5

first two are correct.

Answer 6

yes :)

Answer 7

3.\[\csc x=2,\sin x=\frac{ 1 }{ 2 }=\sin \left( 2n \pi+\frac{ \pi }{ 6 } \right),\]
where n is an integer.
for n=1,what you get?

Answer 8

\[\sin(2\pi+\pi/6)\]

Answer 9

4. also
\[\sin x=\sin \left( 2n \pi+\frac{ 5 \pi }{ 6 } \right),for ~n=1\]
what you get?

Answer 10

i don't really know how to solve or do these :/

Answer 11

\[x=2 \pi+\frac{ \pi }{ 6 }=\frac{ 12 \pi+1 \pi }{ 6 }=?\]

Answer 12

13pi/6

Answer 13

yes you are correct.
I suppose all are correct and you have to verify it.

Answer 14

yes!!

Answer 15

\[2cosx+1=0 \]

Answer 16

1 .
at x=pi/6\[2\sin x-1=2\sin \frac{ \pi }{ 6 }-1=2*\frac{ 1 }{ 2 }-1=1-1=0\]
which is true.

Answer 17

does that equation equal to 2pi/3 and 4pi/3

Answer 18

which question you are talking about?

Answer 19

2cosx+1=0

Answer 20

it is 5th or 6th.

Answer 21

solve equation: 2cosx+1=0, is the answer 2pi/3 and 4pi/3

Answer 22

\[2\cos \frac{ 2 \pi }{ 3 }+1=2\cos \left( \pi-\frac{ \pi }{ 3 } \right)+1=-2\cos \frac{ \pi }{ 3 }+1=-2*\frac{ 1 }{ 2 }+1=0\]
which is true.
also it is true for 4 pi/3

Answer 23

ok :)

Answer 24

at x=2pi/3
\[2\cos ^2x-\cos x-1=2\cos ^2\frac{ 2\pi }{ 3 }-\cos \frac{ 2 \pi }{ 3 }-1=2\cos ^2\left( \pi-\frac{ \pi }{ 3 } \right)-\cos \left( \pi-\frac{ \pi }{ 3 } \right)-1\]
\[=2\left[ -\cos \frac{ \pi }{ 3 } \right]^2-\left[ -\cos \frac{ \pi }{ 3 } \right]-1\]
\[=2\cos ^2\frac{ \pi }{ 3 }+\cos \frac{ \pi }{ 3 }-1\]
\[=2\left( \frac{ 1 }{ 2 } \right)^2+\frac{ 1 }{ 2 }-1\]
\[=\frac{ 1 }{ 2 }+\frac{ 1 }{ 2 }-1=1-1=0\]
which is true.