# Thread: solving for angle in equation

1. ## 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..

2. 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?

3. 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?

4. 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)

5. 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

6. 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

$6\sin ^2(x) + \sin(x) -1 = (3\sin(x)-1)(2\sin(x)+1)=0$

7. 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

8. 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

9. Originally Posted by Archie Meade
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?

10. 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?
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.

11. 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.

12. so i can say that this answer is 19. 47 degrees

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