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# C3 Trig question watch

1. Solve in the interval 0< (theta) < 180 degrees

2cot^2(theta)= 7cosec(theta) - 8

Confused on how to answer this.

I made the 2 cot^2 theta 2cos^2(theta)/sin^2(theta) and the 7cosec(theta) 7/sin(theta) but now I don't know what to do many thanks
2. Multiply through by and then replace the cos with the equivalent in sin.
3. I still confused, So I've times by sin^2 (theta) so I've got

2cos^2(theta)=7sin(theta) - 8 sin^2(theta)

So then 8sin^2(theta)+2cos^2(theta) = 7sin(theta)

so 7sin(theta) = 10 which doesnt work? What have I done wrong?
4. I still confused, So I've times by sin^2 (theta) so I've got

2cos^2(theta)=7sin(theta) - 8 sin^2(theta)

So then 8sin^2(theta)+2cos^2(theta) = 7sin(theta)

so 7sin(theta) = 10 which doesnt work? What have I done wrong?
5. (Original post by joestevens2092)
I still confused, So I've times by sin^2 (theta) so I've got

2cos^2(theta)=7sin(theta) - 8 sin^2(theta)

So then 8sin^2(theta)+2cos^2(theta) = 7sin(theta)

so 7sin(theta) = 10 which doesnt work? What have I done wrong?
Ah, I assume you've tried to use ??

Unfortunately that only works when the coefficients are the same (obviously you'd need to change the constant to match the coefficients).

Re-write the cos^2 in terms of sin and then solve as you should be left with a quadratic in sin.
6. got it cheers

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