Question

Help with solving trigonometric equations of the quadratic type

Answer 1

Find all of the solutions of the equation 3sin(pi/2 + x) = 3 on the interval [0,2π].

Answer 2

\[\sin(\frac{ \pi }{ 2 }+x)=\frac{ 3 }{ 3 }=1\]

Answer 3

@phi I am confused what the next step would be

Answer 4

if the problem were
sin(y)= 1
can you solve for y ?

Answer 5

is this a cofunction identity?

Answer 6

with what you said the last time would you set sin to sin^-1(1)=y?

Answer 7

you do
\[ \sin^{-1} (\sin (y)) = \sin^{-1} 1 \]
the left side with inverse sin of sin "undoes" the sin and you get
\[ y = \sin^{-1} 1\]

Answer 8

if
sin(y) = 1
and
sin(pi/2 + x) = 1
then
sin(y) = sin(pi/2 + x)
apply the inverse sin to both sides (to undo the sin). to get
y = pi/2 + x
and so
x= y - pi/2

Answer 9

so for this you would do: \[\sin ^{-1}(\sin(\frac{ \pi }{ 2 }+x))=\sin ^{-1}\frac{ \pi }{ 2 }

And it would ened

Answer 10

end* to be

Answer 11

omy god i messed it up T_T

Answer 12

\[\sin ^{-1}(\sin(\frac{ \pi }{ 2 }+x))=\sin ^{-1}\frac{ \pi }{ 2 }\]

Answer 13

\[x=\sin ^{-1}\frac{ \pi }{ 2 }\]

Answer 14

you lost the bubble.
the problem is
\[ \sin(\frac{ \pi }{ 2 }+x)= 1 \\
\sin ^{-1}(\sin(\frac{ \pi }{ 2 }+x))=\sin ^{-1}1 \]

Answer 15

the left side simplifies to pi/2 +x
the right side is asking for the angle whose sin is 1

Answer 16

I lost the 1 T_T

Answer 17

but hopefully you know sin 90 = 1
or in radians, sin pi/2 = 1
so pi/2 = sin^-1 1

Answer 18

so basically it is saying \[\frac{ \pi }{ 2 }=\frac{ \pi }{2 }\]

Answer 19

it's saying
\[ \frac{\pi}{2} + x = \frac{\pi}{2} \]
which means x=0

Answer 20

Question, What does it mean by, on the interval of [0,2pi]?

Answer 21

sin is defined for all angles, but it repeats after 360 degrees (or 2 pi radians)
so there are an infinite number of solutions to your problem: x= 2pi k for any integer k
in this case, they are restricting the answers to the first 2pi radians

the [ in [0,2pi] means the limit is included (as opposed to (0,2pi) where the parens mean the 0 and 2pi are not in the interval)

looking at that interval , the answer would be x=0 and x=2pi

Answer 22

Okay, that makes a lot more sense.

Answer 23

Thank you so much

Answer 24

I think I can possibly try these on my own now

Answer 25

Question though, What do you do if their are two right next to eachother for example:

\[\sec(x)\csc(x)=2\csc(x)\]

Answer 26

i know you would have to use sec=1/cos(x) and csc(x)=1/sin(x)

Answer 27

and 2(1/sin(x)

Answer 28

would you flip them after so it looks like \[\cos(x)\sin(x)=\frac{ 1 }{ 2 }\sin(x)\]

Answer 29

@phi

Answer 30

yes, exactly

Answer 31

Wouldn't the left side be an identity? One moment as i try to find it

Answer 32

Never mind was thinking of cos^2(x)+sin^2(x)=1

Answer 33

to solve, I am tempted to "cancel" i.e. divide both sides by sin x
but that will leave out some of the answer.
the way to solve is add -1/2 sin x to both sides to get
\[ \cos(x)\sin(x)-\frac{ 1 }{ 2 }\sin(x)=0 \]
factor out sin x
\[ (\cos(x)-\frac{ 1 }{ 2 })\sin(x) =0 \]

Answer 34

now we have two factors, and if *either* is zero, we will get 0 when we multiply
in other words, we solve for
cos x - 1/2 =0
and for
sin x = 0

Answer 35

\[\cos(x)=\frac{ 1 }{ 2 }\]

Answer 36

\[x=\cos ^{-1}\frac{ 1 }{ 2 }\]

Answer 37

\[x=\frac{ \pi }{3 }\]

Answer 38

and for sin \[x=\sin ^{-1}0\]

Answer 39

x=0

Answer 40

what interval are you solving ?
there are other soln's for cos = 1/2

Answer 41

\[[0,2\pi]\]