nt just answers-how to solve: 1) tan^2 x - sq rt(3) tan x = 0?

Question

2) sq rt(3)cot xsinx + 2cos^2x=0
3)cosxtanx-sin^2x=0
4)cosx-2cosxsinx=0
5)sin 2x=cos x

Answer 1

1. tan^2 x - sq rt(3) tan x = 0
(tanx)(tanx) = sq rt(3) tan x
so if you cancel out tan x on each side youre left with,

tanx = sq rt(3)
dont forget the tanx you cancelled out. You cant just leave that be:
tanx = 0
so now that you have there 2 equations, you can solve for x and get the values for which tanx = 0 and sq rt(3)

x = 0 deg 60 deg 180 deg 240deg



2. question 2. is similar to question 1. firstly, cotx = cosx/sinx.
So cotx sinx = cosx/sinx * sinx = cosx
So the whole equation would get simplified to:
sq rt(3)cosx + 2cos^2x=0

This is now exactly the same as question 1. Try this yourself. The only way to truly know and understand math and grasp the concept is to do it.



3. tanx = sinx/cosx
So, cosxtanx = cosx * sinx/cosx = sinx
So the whole equation is reduced to sinx - sin^2x=0

Now this is again exactly the same as question 1 and 2. So again do it yourself and grasp the concept.



4. I'll do this one for you so you can double check 2. and 3.
cosx - 2cosxsinx = 0
transfer 2cosxsinx over:
cos x = 2cosxsinx
cancel cos x on each side:
1 =- 2sinx
bring the 2 over:
1/2 = sinx
this is your first equation. from this solve for x when sinx = 1/2:
x = 30 deg, 150 deg.

dont forget the cosx you cancelled out:
cos x = 0
x = 90 deg 270deg.

so the values for x are 30deg 90deg 150deg 270deg



5. sin 2x = cos x

since, sin2x = 2(sinxcosx)
2(sinxcosx) = cos x

cancel the cos x:

2sinx = 1
sinx = 1/2
not forgetting the cosx = 0

solve for x again.


Trig requires a whole lot of practice for you to grasp your concept. Thats probably why most of these sums are actually almost the same requiring almost the same step. I suggest practising these sums with your text or notes with you so that you can refer and find out what your next step is. It's easy once you get the hang of it and really interesting.
Dont let math beat you!

Answer 2

1.tan^2x-3^1/2tanx=0 =>tanx(tanx-3^1/2)=0 =>tanx=0 or tanx=3^1/2
if tanx=0 ,x= n*pi if tanx=3^1/2 ,x= n*pi+tan inverse3^1/2
2.3^1/2cotxsinx+2cos^2x=0 =>cosx(3^1/2+2cosx)=0
=>cosx=0 or x= (2n+1)*pi/2 cosx=-3^1/2/2 or x=2n*pi plus or minus pi/6
3.cosxtanx-sin^2x=0 =>sinx(1-sinx)=0=>sinx=0 or sinx=1
if sin x=0,x= n*pi if sinx=1 x=n pi+(-1)^n*pi/2
4.cosx-2cosxsinx=0=>cosx(1-2sinx)=0=>cosx=0 or sinx=1/2
if cosx=0 x= (2n+1)*pi/2 if sinx=1/2 x=(2n+1)*pi/2
5.sin2x=cosx=>2sinxcosx=cosx=>cosx(2sinx-1)=0
cosx=o so x= (2n+1)*pi/2 sinx=1/2 sox=(2n+1)*pi/2
in all the above cases i have given the general solutions
in all the above cases n is an element of integers and for different integral values of n you will get different solutions satisfying the equations

Answer 3

1. factor tanx
tanx (tanx - sqrt(3) = 0
now set each factor to 0
so a) tanx = 0 or b) tanx - sqrt(3) =
x= 0 tanx=sqrt(3) do inverse of tan
x = 60

4. factor cosx
cosx (1-2sinx) = 0
so a) cosx = 0 or b) 1 - 2sinx=0
x-90 1=2sinx
sinx=1/2
x=30

Answer 4

is it for any value of x or is there a range

if it is for any value:

3:cosxtanx - sin^2x = 0
= cosx(sinx/cosx) - sin^2x = 0
= sinx(1-sinx) = 0
= either sinx = 0 or (1-sinx) = 0
could be 0 or 90 degrees

Answer 5

There are other answers. Remember for question 1 that tan is positive in the 3rd quadrant so there is another answer there.(240)
Similarly in question 5. Sin is positive in quadrant 2.(150)

The 4 quadrants are All, sin, tan, cos. you need to know this and check all quadrants for alternative answers

Answer 6

assuming i read you correctly

1.)
(tan(x))^2 - sqrt(3)tanx = 0
(tan(x))(tan(x) - sqrt(3)) = 0

tan(x) = 0
x = 0° or 180°

tan(x) - sqrt(3) = 0
tan(x) = sqrt(3)
x = tan^-1(sqrt(3))
x = 60°

ANS : x = 0°, 60°, 180°

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2.)
(sqrt(3))(cot(x))(sin(x)) + 2(cos(x))^2 = 0
sqrt(3)(cos(x)/sin(x))sin(x) + 2(cos(x))^2 = 0
sqrt(3)((cos(x)sin(x))/sin(x)) + 2(cos(x))^2 = 0
sqrt(3)cos(x) + 2(cos(x))^2 = 0
2(cos(x))^2 + sqrt(3)cos(x) = 0
(cos(x))(2cos(x) + sqrt(3)) = 0

cos(x) = 0
x = 90° or 270°

2cos(x) + sqrt(3) = 0
2cos(x) = -sqrt(3)
cos(x) = (-sqrt(3))/2
x = 150° or 210

ANS : x = 90°, 150°, 210°, 270°

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3.)
(cos(x))(tan(x)) - (sin(x))^2 = 0
(cos(x) * (sin(x)/cos(x))) - (sin(x))^2 = 0
((cos(x)sin(x))/cos(x)) - (sin(x))^2 = 0
sin(x) - (sin(x))^2 = 0
(sin(x))(1 - sin(x) = 0

sin(x) = 0
x = 0° or 180°

1 - sin(x) = 0
-sin(x) = -1
sin(x) = 1
x = 90°

ANS : 0°, 90°, or 180°

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4.)
cos(x) - 2(cos(x)sin(x)) = 0
(cos(x))(1 - 2sin(x)) = 0

cos(x) = 0
x = 90° or 270°

1 - 2sin(x) = 0
-2sin(x) = -1
sin(x) = (1/2)
x = 30°

ANS : x = 30°, 90°, 270°

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5.)
sin(2x) = cos(x)
sin(2x) = 2(sin(x)cos(x))

2(sin(x)cos(x) = cos(x)
2(sin(x)cos(x)) - cos(x) = 0
(cos(x))(2sin(x) - 1) = 0

cos(x) = 0
x = 90°, 270°

2sin(x) - 1 = 0
2sin(x) = 1
sin(x) = (1/2)
x = 30°

ANS : x = 30°, 90°, or 270°

Answer 7

no idea