[Jozzers]

Q. Solve the following equation for :

So far I have:

I don't know where to go from there on.

Thanks in advance for any help.

[MHRed]

You may be overcomplicating multiply through by sin^2 or cos^2 at the beginning, this will give you just 1 trig function.

If I've been ambiguous I apologise, and I'll explain further

[ztibor]

(Original post by Jozzers)
Q. Solve the following equation for :

So far I have:

I don't know where to go from there on.

Thanks in advance for any help.

Use

at the LHS so

[f1mad]

At the beginning, multiply through by sin^2theta following on from MHRed. It leaves a simple quadratic.

[raheem94]

(Original post by Jozzers)
Q. Solve the following equation for :

So far I have:

I don't know where to go from there on.

Thanks in advance for any help.

Multiply both sides by

[PeteyB26]

cosec^2(x) = 9sec^2(x)

convert cosec onto 1/sin and sec into 1/cos

1/sin^2(x) = 9/cos^2(x)

convert cos^2 into 1-sin^2

1/sin^2(x) = 9/1-sin^2(x)

cross multiply

1-sin^2(x) = 9sin^2(x)

add sin^2

1 = 10sin^2(x)

divide by 10

1/10 = sin^2(x)

squareroot

root(1/10) = sin(x)

inverse sin

sin^-1 (root(1/10)) = x

then take 180 minus your value for second principal value.

...hope you can follow this

[PeteyB26]

Hahahaha, when you spend ages writing a reply and someone else does it much simpler, cool move Pete xD

[James A]

(Original post by PeteyB26)
cosec^2(x) = 9sec^2(x)

convert cosec onto 1/sin and sec into 1/cos

1/sin^2(x) = 9/cos^2(x)

convert cos^2 into 1-sin^2

1/sin^2(x) = 9/1-sin^2(x)

cross multiply

1-sin^2(x) = 9sin^2(x)

add sin^2

1 = 10sin^2(x)

divide by 10

1/10 = sin^2(x)

squareroot

root(1/10) = sin(x)

inverse sin

sin^-1 (root(1/10)) = x

then take 180 minus your value for second principal value.

...hope you can follow this

this is exactly what i was thinking. Just rewriting cosec^2 and sec^2 is the big trick in this question!

I don't know why people are going for the tan's and cot^s lol

[PeteyB26]

(Original post by James A)
this is exactly what i was thinking. Just rewriting cosec^2 and sec^2 is the big trick in this question!

I don't know why people are going for the tan's and cot^s lol

Absolutely, that's the problem with trig identities though, if you go down the wrong path at first you can spend ages just getting yourself in a muddle! They used to scare me so much in C2.. Hahaha

[GreenLantern1]

(Original post by PeteyB26)
cosec^2(x) = 9sec^2(x)

convert cosec onto 1/sin and sec into 1/cos

1/sin^2(x) = 9/cos^2(x)

convert cos^2 into 1-sin^2

1/sin^2(x) = 9/1-sin^2(x)

cross multiply

1-sin^2(x) = 9sin^2(x)

add sin^2

1 = 10sin^2(x)

divide by 10

1/10 = sin^2(x)

squareroot

root(1/10) = sin(x)

inverse sin

sin^-1 (root(1/10)) = x

then take 180 minus your value for second principal value.

...hope you can follow this

Or you can do an even simpler method.

convert cosec into 1/sin and sec into 1/cos

1/sin^2(x) = 9/cos^2(x)

Cross multiply and you get:

cos^2(x)/sin^2(x)=9

1/tan^2(x)=9

Reciprocate it

tan^2(x)=1/9

tan(x)=1/3

x=18.43,198.43

[just george]

or












edit: typically someone beats me to it

[James A]

(Original post by PeteyB26)
Absolutely, that's the problem with trig identities though, if you go down the wrong path at first you can spend ages just getting yourself in a muddle! They used to scare me so much in C2.. Hahaha

yeah i know, it helps alot if i had a brain that could see five steps in advance haha

[Jozzers]

Ohhh I get it now.

Thank you everyone!