# solving for angle in equation

• Apr 14th 2010, 08:31 AM
sigma1
solving for angle in equation
am trying to find the value for x between 0 and 360 degress. but i think am doing something wrong.

$2-sinx = cos^2x + 7sin^2x$

$2-sinx = 1-sin^2x + 7sin^2x$

$2-sinx = 1- 6sin^2x$

can someone help me to solve this am stuck here..
• Apr 14th 2010, 08:39 AM
Sudharaka
Quote:

Originally Posted by sigma1
am trying to find the value for x between 0 and 360 degress. but i think am doing something wrong.

$2-sinx = cos^2x + 7sin^2x$

$2-sinx = 1-sin^2x + 7sin^2x$

$\color{red}{2-sinx = 1- 6sin^2x}$

can someone help me to solve this am stuck here..

Dear sigma1,

There is a mistake in the highlighted line. Can you find it?
• Apr 14th 2010, 09:17 AM
sigma1
Quote:

Originally Posted by Sudharaka
Dear sigma1,

There is a mistake in the highlighted line. Can you find it?

yes i see the error.

it should be

$2-sinx = 1+6sin^2x$
$1-sinx = 6sin^2x$
$-sinx -6sin^2x=-1$
$-sinx(1+6sinx)=-1$

where do i go from here?
• Apr 14th 2010, 09:54 AM
harish21
Quote:

Originally Posted by sigma1
am trying to find the value for x between 0 and 360 degress. but i think am doing something wrong.

$2-sinx = cos^2x + 7sin^2x$

$2-sinx = 1-sin^2x + 7sin^2x$

$2-sinx = 1 + 6sin^2x$

can someone help me to solve this am stuck here..

So you will have:

$6sin^2x + sinx -1 =0$

let $u = sinx$ and you can see that this is a quadratic equation:

$6u^2 +u -1 = 0$

Now solve for $u$ using the quadratic equation formula, which will give you the value of $u$, that is sinx.

after getiing $sinx$, you can calculate the angle(x)
• Apr 14th 2010, 10:08 AM
harish21
Quote:

Originally Posted by sigma1
yes i see the error.

it should be

$2-sinx = 1+6sin^2x$
$1-sinx = 6sin^2x$
$-sinx -6sin^2x=-1$
$-sinx(1+6sinx)=-1$

where do i go from here?

this is not going to work. refer to the above post and work on the quadratic equation
• Apr 14th 2010, 10:28 AM
e^(i*pi)
Quote:

Originally Posted by harish21
So you will have:

$6sin^2x + sinx -1 =0$

let $u = sinx$ and you can see that this is a quadratic equation:

$6u^2 +u -1 = 0$

Now solve for $u$ using the quadratic equation formula, which will give you the value of $u$, that is sinx.

after getiing $sinx$, you can calculate the angle(x)

This equation will also factorise

$6\sin ^2(x) + \sin(x) -1 = (3\sin(x)-1)(2\sin(x)+1)=0$
• Apr 14th 2010, 10:46 AM
sigma1
Quote:

Originally Posted by harish21
this is not going to work. refer to the above post and work on the quadratic equation

ok thanks for that. i solved the equation and got 2 values for u.

u =1/3 and u = - 1/2

which one of the values should i use.

i used -1/2 though to get $6sin^ -[\frac{1}{2}] - 1 = 0$

sin^2x = \frac{1}{2}
sinx = 1/2
x = $sin^{-1} 1/2$
x = 30
• Apr 14th 2010, 10:56 AM
Hi sigma1,

the quadratic equation allowed you to discover the 2 possible values for $sinx$

Now you need to find the valid values of the angle x.

$sinx$ may be conveniently thought of as the vertical co-ordinate
of a point on a circle centred at (0,0) with radius 1.

Then you can see, using a sketch that a vertical co-ordinate of $\frac{1}{3}$

gives 2 possible angles, $x=sin^{-1}\left(\frac{1}{3}\right)$

and $180^o-sin^{-1}\left(\frac{1}{3}\right)$

For $sinx=-\frac{1}{2}$ on the vertical axis,

you can calculate the acute angle $sin^{-1}\left(\frac{1}{2}\right)$

to get the acute angle the point on the circle makes under the x-axis.

Then subtract this from 360 degrees and add it to 180 degrees.

You really need to understand that procedure
• Apr 14th 2010, 12:44 PM
sigma1
Quote:

Hi sigma1,

the quadratic equation allowed you to discover the 2 possible values for $sinx$

Now you need to find the valid values of the angle x.

$sinx$ may be conveniently thought of as the vertical co-ordinate
of a point on a circle centred at (0,0) with radius 1.

Then you can see, using a sketch that a vertical co-ordinate of $\frac{1}{3}$

gives 2 possible angles, $x=sin^{-1}\left(\frac{1}{3}\right)$

and $180^o-sin^{-1}\left(\frac{1}{3}\right)$

For $sinx=-\frac{1}{2}$ on the vertical axis,

you can calculate the acute angle $sin^{-1}\left(\frac{1}{2}\right)$

to get the acute angle the point on the circle makes under the x-axis.

Then subtract this from 360 degrees and add it to 180 degrees.

You really need to understand that procedure

i am a bit unclear on the theory you trying to bring across to me... its the first am seeing something like this... could i have missed this in class(Surprised)?
• Apr 14th 2010, 12:55 PM
Quote:

Originally Posted by sigma1
i am a bit unclear on the theory you trying to bring across to me... its the first am seeing something like this... could i have missed this in class(Surprised)?

quite possibly,
it may not always be brought across as

$sin(angle)=y\ co-ordinate$

$cos(angle)=x\ co-ordinate$

in a circle, centre (0,0) and radius 1

However, i feel it's quite easy to work with that way.

You may instead be presented with

sin is + in the 1st and 2nd quadrants, sin is - in the 3rd and 4th.
cos is + in the 1st and 4th quadrants, cos is - in the 2nd and 3rd.

this corresponds to one angle in the 1st quadrant, 0 to 90 degrees
and another angle in the 2nd quadrant, 90 to 180 degrees.

Then work from there knowing that the angle in the 2nd quadrant is 180-(angle in 1st)

Similarly for cos, using the guidelines for that.

i much prefer the horizontal and vertical co-ordinates way,
after all that's why we can give sin and cos their polarities in the 4 quadrants.

Whichever way you do it, remember sinx points out 2 angles as does cosx.
• Apr 14th 2010, 01:14 PM
harish21
Quote:

Originally Posted by sigma1
ok thanks for that. i solved the equation and got 2 values for u.

u =1/3 and u = - 1/2

which one of the values should i use.

i used -1/2 though to get $6sin^ -[\frac{1}{2}] - 1 = 0$

sin^2x = \frac{1}{2}
sinx = 1/2
x = $sin^{-1} 1/2$
x = 30

So your values are $u = \frac{1}{3}$ and $u =\frac{-1}{2}$, which you got correct.

That means:

$sinx = \frac{1}{3}$
or
$sinx=\frac{-1}{2}$

now calculate x.

Note that your question is asking for x values between 0 to 360 degrees.
• Apr 14th 2010, 01:53 PM
sigma1
so i can say that this answer is 19. 47 degrees
• Apr 14th 2010, 02:01 PM
harish21
Quote:

Originally Posted by sigma1
so i can say that this answer is 19. 47 degrees

yes