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# C3: Solving Trigonometric Equations Watch

1. 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.
2. 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
3. (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
4. At the beginning, multiply through by sin^2theta following on from MHRed. It leaves a simple quadratic.
5. (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
6. 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)

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.

7. Hahahaha, when you spend ages writing a reply and someone else does it much simpler, cool move Pete xD
8. (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)

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.

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
9. (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
10. (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)

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.

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
11. or

edit: typically someone beats me to it
12. (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
13. Ohhh I get it now.

Thank you everyone!

Updated: April 3, 2012
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