Trig questions (1 Viewer)

[od9895]

Hi
Could anyone help me with these trig questions

Find general solutions?

1. 2 sin^2x + cosx =1
2. 4 +sin a = 6cos^2 a
3. 1 + 2cos^2x= 5sinx
4. tanx =sinx
5. sin2a= tana
6. sin3x =sin2x

THanks

 

[Aesytic]

for questions 1-3 consider the identity sin^2 x + cos^2 x = 1
for 4 change tan to sin/cos
for 5 use double angle formula for sin2a and change tan to sin/cos
for 6 double angle formula and consider sin(2x + x) instead of sin3x [unless you've learnt the identity for sin3x off by heart]

 

[Carrotsticks]

Question 1 :

Question 4:

Question 5:

Questions 2 and 3 use similar methods to question 1.

Question 6 is just very long, and I need to head off soon.

 

[RivalryofTroll]

6. sin3x =sin2x

sin3x = sin2x
sin(2x+x) = 2sinxcosx
sin2x.cosx + sinx.cos2x = 2sinxcosx
2sinxcosx.cosx + sinx.(2cos^2x -1) = 2sinxcosx
2sinxcos^2x + 2sinxcos^2x - sinx = 2sinxcosx
4sinxcos^2x - 2sinxcosx - sinx = 0
sinx(4cos^2x - 2cosx - 1) = 0
sinx = 0
x = 0, 180, 360
4cos^2x - 2cosx - 1 = 0
Let x = cosx
4x^2 - 2x - 1 = 0
x = (1+/- root 5)/4

cos x = (1 + root5)/4
x = 36 and 360 - 36
cos x = (1 - root5)/4
x = 180-108 and 180 + 108

x = 36 ,324 AND x = 72, 288

Therefore x = 0, 36, 72, 180, 288, 324, 360

(I think I got it wrong, my figures look bad)

 

[Carrotsticks]

Oh and I made a mistake in my solutions.

It says:



It should actually be

 

[SpiralFlex]

Oh and I made a mistake in my solutions.

It says:



It should actually be

What tablet do you use?

 

[Carrotsticks]

What tablet do you use?

Wacom Bamboo Fun

Cost me about $200 from eBay.

 

[bleakarcher]

I got x=0,2pi,pi/5,3pi/5,pi,7pi/5,9pi/5 which equate to x=0,360,36,108,180,252,324 in degrees

 

[RivalryofTroll]

I got x=0,2pi,pi/5,3pi/5,pi,7pi/5,9pi/5 which equate to x=0,360,36,108,180,252,324 in degrees

Hey bleakarcher, can you please look at my solution ^up there and tell me where I went wrong?
I'm assuming that you got it right and I got it wrong since you got the most obvious figures like I did and a few different figures.

EDIT:
My figures were, if you can't be bothered scrolling up for the working out,:
x = 0, 36, 72, 180, 288, 324, 360

 

[bleakarcher]

sin3x = sin2x
sin(2x+x) = 2sinxcosx
sin2x.cosx + sinx.cos2x = 2sinxcosx
2sinxcosx.cosx + sinx.(2cos^2x -1) = 2sinxcosx
2sinxcos^2x + 2sinxcos^2x - sinx = 2sinxcosx
4sinxcos^2x - 2sinxcosx - sinx = 0
sinx(4cos^2x - 2cosx - 1) = 0
sinx = 0
x = 0, 180, 360
4cos^2x - 2cosx - 1 = 0
Let x = cosx
4x^2 - 2x - 1 = 0
x = (1+/- root 5)/4

cos x = (1 + root5)/4
x = 36 and 360 - 36
cos x = (1 - root5)/4
x = 180-108 and 180 + 108

x = 36 ,324 AND x = 72, 288

Therefore x = 0, 36, 72, 180, 288, 324, 360

(I think I got it wrong, my figures look bad)

That?

 

[SpiralFlex]

Wacom Bamboo Fun

Cost me about $200 from eBay.

I remember those! I want one after the HSC! After I get a job.

 

[RivalryofTroll]

That?

cos x = (1 - root5)/4
x = 108
Therefore, x = 180 - 108 and 180 + 108

Cos is negative in the 2nd and 3rd Quadrants?
A - (x)
S - Cos Negative (180 - x)
T - Cos Negative (180 + x)
C - (360 - x)

 

[Aesytic]

108 is in the 2nd quadrant already, but doing 180-108, you get 72, which is in the first quadrant and doesn't make sense since [1-root5]/4 is a negative value. what you could do [what i usually do but i don't know how others do it] is change the value of [1-root5]/4 to make it positive instead, then finding inverse cos of that to get 72 before doing 180-x and 180+x