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C3: Solving Trigonometric Equations

Q. Solve the following equation for

0^o \leq \theta \leq 360^o

\mathrm{cosec^2} \theta = 9\sec^2 \theta

So far I have:

1+\cot^2 \theta = 9(1+\tan^2 \theta)

I don't know where to go from there on. :s-smilie:

Thanks in advance for any help.

Reply 1

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 :smile:

Reply 2

ztibor

Original post by Jozzers

Q. Solve the following equation for

0^o \leq \theta \leq 360^o

\mathrm{cosec^2} \theta = 9\sec^2 \theta

So far I have:

1+\cot^2 \theta = 9(1+\tan^2 \theta)

I don't know where to go from there on. :s-smilie:

Thanks in advance for any help.

Use

\displaystyle cot \theta =\frac{1}{tan \theta}

at the LHS so

\displaystyle \left (1+tan^2\theta \right )\left( \frac{1}{tan^2 \theta}-9\right )=0

(edited 12 years ago)

Reply 3

f1mad

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

Reply 4

raheem94

Original post by Jozzers

Q. Solve the following equation for

0^o \leq \theta \leq 360^o

\mathrm{cosec^2} \theta = 9\sec^2 \theta

So far I have:

1+\cot^2 \theta = 9(1+\tan^2 \theta)

I don't know where to go from there on. :s-smilie:

Thanks in advance for any help.

\displaystyle \mathrm{cosec^2} \theta = 9\sec^2 \theta

\displaystyle \frac{1}{sin^2\theta} = \frac9{cos^2\theta}

Multiply both sides by

\displaystyle sin^2\theta

Reply 5

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 :smile:

Reply 6

PeteyB26

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

Reply 7

F1's Finest

Original post by PeteyB26

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

Reply 8

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

Reply 9

GreenLantern1

Original post by PeteyB26

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

Reply 10

just george

cosec^2\theta = 9sec^2\theta

\frac{1}{sin^2\theta} = \frac{9}{cos^2\theta}

1 = \frac{9sin^2\theta}{cos^2\theta}

1 = 9tan^2\theta

\frac{1}{9} = tan^2\theta

\pm(\frac{1}{3}) = tan\theta

edit: typically someone beats me to it :biggrin:

(edited 12 years ago)

Reply 11

F1's Finest

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