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# C2 trig question watch

Solve the following for [theta] in the interval 0 < [theta] < (and equal to)360

tan[theta] = tan[theta](2 + 3sin[theta])

Is it possible to do:

tan[theta] / tan[theta] = 2 + 3sin[theta]
1 = 2 + 3sin[theta]
sin[theta] = -1/3
[theta] = -19.5

This gives me the values of 199.5 and 340.5, but the book also gives the values 180 and 360.

Why am I missing these values? I don't understand? Could someone kindly explain to me where I'm going wrong?

Thanks
2. Dividing through by variables can lose you solutions. Instead of cancelling, arrange into a quadratic and factorise.

Suppose you had x = x(2+3x). Clearly this works if x = 0, but dividing through by x just leaves 1 = 2 + 3x so x = -1/3 (and you miss you out the solution x = 0)
3. Right, I understand tan[theta] = 0 is also a solution (so theta = 180, 360), but I'm not sure how to rearrange to prove this... sorry my mind has just gone blank
4. Move the tan theta on the left to the right:

0 = tan theta (2 + 3 sin theta) - tan theta

Expand brackets

Then factorise.
5. (Original post by mya369)
Move the tan theta on the left to the right:

0 = tan theta (2 + 3 sin theta) - tan theta

Expand brackets

Then factorise.
Got it, thanks a lot

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