Solving the Equation cos (x)=-0.8

[fordy2707]

Hi all,just wondering if someone could cast an eye over this,

find all the solutions to the following equation at intervals 0,6$\pi$

cos (x)= -0.8

= $143^{\circ}$

my answer by using the quadrant chart

= 180$^{\circ}$-143$^{\circ}$= $37^{\circ}$

= 180$^{\circ}$ + 37$^{\circ}$ = 217$^{\circ}$

so

= $143^{\circ}$ x $\pi$ / 180

= 143$\pi$ / 180
+2$\pi$= 503$\pi$ / 180
+2$\pi$ = 863$\pi$ / 180

and the 2nd set of solutions

=217$^{\circ}$ x $\pi$ / 180

=217 $\pi$ / 180
+2$\pi$=577 $\pi$ /180
+2$\pi$ =937$\pi$ /180

I was wondering about my answers as the fraction figures seem very large

 

[MarkFL]

I would look at the symmetry of the solutions about $x=2k\pi$ where $k\in\mathbb{Z}$ to write:

\(\displaystyle x=2k\pi\pm\arccos\left(-\frac{4}{5}\right)\)

And so on the given interval, we find:

\(\displaystyle x=0+\arccos\left(-\frac{4}{5}\right)\approx2.498091544796509\)

\(\displaystyle x=2\pi-\arccos\left(-\frac{4}{5}\right)\approx3.7850937623830774\)

\(\displaystyle x=2\pi+\arccos\left(-\frac{4}{5}\right)\approx8.781276851976095\)

\(\displaystyle x=4\pi-\arccos\left(-\frac{4}{5}\right)\approx10.068279069562664\)

\(\displaystyle x=4\pi+\arccos\left(-\frac{4}{5}\right)\approx15.064462159155681\)

\(\displaystyle x=6\pi-\arccos\left(-\frac{4}{5}\right)\approx16.35146437674225\)

 

[HallsofIvy]

Hi all,just wondering if someone could cast an eye over this,

find all the solutions to the following equation at intervals 0,6$\pi$

cos (x)= -0.8

= $143^{\circ}$

To begin with this says that "cos(x)= $143^{\circ}$" which is nonsense. You mean to say "x= $143^{\circ}$" although I would say that was incorrect because x is clearly supposed to be in radians, not degrees. And, of course, "143 degrees" is not correct because it does not include the decimal part. x is approximately $143^{\circ}$ though that is not a very good approximation.

[/tex]my answer by using the quadrant chart

= 180$^{\circ}$-143$^{\circ}$= $37^{\circ}$

= 180$^{\circ}$ + 37$^{\circ}$ = 217$^{\circ}$

so

= $143^{\circ}$ x $\pi$ / 180

= 143$\pi$ / 180
+2$\pi$= 503$\pi$ / 180
+2$\pi$ = 863$\pi$ / 180

and the 2nd set of solutions

=217$^{\circ}$ x $\pi$ / 180

=217 $\pi$ / 180
+2$\pi$=577 $\pi$ /180
+2$\pi$ =937$\pi$ /180

I was wondering about my answers as the fraction figures seem very large

937/180= 5.2055... which is less than 6 and you were asked about values less that $6\pi$.

However, I would have set my calculator to radian measure to begin with. Is there any reason you took this "detour" through degrees? You would get a much more accurate answer since you just ignored the decimal part of the degree measure.

 

[fordy2707]

To begin with this says that "cos(x)= $143^{\circ}$" which is nonsense. You mean to say "x= $143^{\circ}$" although I would say that was incorrect because x is clearly supposed to be in radians, not degrees. And, of course, "143 degrees" is not correct because it does not include the decimal part. x is approximately $143^{\circ}$ though that is not a very good approximation.937/180= 5.2055... which is less than 6 and you were asked about values less that $6\pi$.

However, I would have set my calculator to radian measure to begin with. Is there any reason you took this "detour" through degrees? You would get a much more accurate answer since you just ignored the decimal part of the degree measure.

I see ,its just the way I was shown ,

I thought 143.1301024 would be rounded up to 143 degrees is this not correct ?

$\frac{143\pi}{180}$ = 2.49582083

and 143.1301024 x $\pi$ = 449.6564782 and when divided by 180 to get radians =2.498091546 would It not be deemed at the same answer when rounded up to 2.5 for example

can you show me how you would have gone about this ? thanks

 

[skeeter]

can you show me how you would have gone about this ? thanks

first, it helps to visualize the two angles between $0$ and $2\pi$ where the cosine value is $-0.8$

Note that angles with a negative value for cosine are in quadrants II and III. In the attached diagram, the terminal sides of angles $\theta$ and $\phi$ intersect the unit circle at $x = -0.8$

The inverse cosine function on your calculator will yield angle values in quadrant I for positive values of cosine, and in quadrant II for negative values of cosine. So, if you type $\cos^{-1}(-0.8)$ into your calculator (in radian mode), the calculator will output $\approx 2.4981$ radians, which is the radian measure for angle $\theta$ in the diagram. You may want to use the "store" feature of your calculator to keep the more accurate approximate value.

To get angle $\phi$ in quadrant III, calculate $2\pi - \cos^{-1}(-0.8) \approx 3.7851$ radians

Now all you have to do is add $2\pi$ and $4\pi$ to both $\theta$ and $\phi$ to get all six angles in the interval $(0,6\pi)$ where the cosine's value is $-0.8$View attachment 5953