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# A2 Trigonometry help?? watch

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1. Hey, can anyonle help me on these questions?
-180< x <180

1) sec^2x + tan^2x = 6

and i got... tan^2x + 1 + tan^2x = 6
and I can't get any further

2) 4cos^2x + 5sinex = 3

I have noo idea how to start this one

3) tanx + cotx = 2

dont know what to do here either
2. For Q1, you know 2tan²x + 1 = 6. So tan²x = 5/2. So tan x = sqrt(5/2). Essentially you just solve for tan x as you would if it were just 'x' or 'y' or whatever .

For Q2, write cos²x = 1 - sin²x, then you have a quadratic equation in sin x . Again, just treat 'sin x' as if it were 'x' or 'y'.

For Q3, just multiply both sides by tan x .
3. (Original post by Simba)
For Q1, you know 2tan²x + 1 = 6. So tan²x = 5/2. So tan x = sqrt(5/2). Essentially you just solve for tan x as you would if it were just 'x' or 'y' or whatever .

For Q2, write cos²x = 1 - sin²x, then you have a quadratic equation in sin x . Again, just treat 'sin x' as if it were 'x' or 'y'.

For Q3, just multiply both sides by tan x .
I would add to Simba's post that you need to consider there are two cases to consider from tan²x = 5/2 and a total of four solutions in that interval.
4. (Original post by Simba)
For Q1, you know 2tan²x + 1 = 6. So tan²x = 5/2. So tan x = sqrt(5/2). Essentially you just solve for tan x as you would if it were just 'x' or 'y' or whatever .

For Q2, write cos²x = 1 - sin²x, then you have a quadratic equation in sin x . Again, just treat 'sin x' as if it were 'x' or 'y'.

For Q3, just multiply both sides by tan x .
AAH thanks for the help...I've just completed them - its annoying to think that I keep forgetting how to do the simple things to them...like multiplying both sides by tan x

Thanks a lot

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