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Trignomertirical Equations Problem

so I'm having problems answering this question:

Solve, for –180° X < 180°

(1 + tan X)(5 sin X 2) = 0
Tanx = sinx/cosx
Then multiply out
Reply 3
What happens if tanx=1 \tan x =-1 and/or sinx=25 \sin x =\frac{2}{5} .

So we have tanx=1x=arctan(1) \tan x = -1 \Rightarrow x=\arctan(-1) <--- standard angle.

Also sinx=25x=arcsin(25) \sin x= \frac{2}{5} \Rightarrow x= \arcsin \left ( \frac{2}{5} \right ) .
Use the fact that tan has a period of π \pi and sin has a period of 2π 2\pi to find all the answers in the specified range.
The question is almost solved for you, there is no manipulation required all you need to do is what I did above.
(edited 8 years ago)
Might also have to use sin^2 + cos^2 = 1 and rearranged
Do the inverse as there are solutions for tan, cos and sin between -1 and 1
Original post by audreygal
Then multiply out


this is not necessary...

if A x B = 0

then either A = 0 or B = 0

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