# trigonometric inequality

• Oct 2nd 2009, 03:22 AM
thereddevils
trigonometric inequality
Find the value of x , valid between -pi and pi , for the following inequality . (I am not allowed to do these using the graphical method)

(1) tan x < cos x

(2) 2 cos 2x > 3cos x -1

I just do not know the steps .The range of x makes it worse . (Headbang) THanks for your time .
• Oct 2nd 2009, 04:40 AM
mr fantastic
Quote:

Originally Posted by thereddevils
Find the value of x , valid between -pi and pi , for the following inequality . (I am not allowed to do these using the graphical method)

(1) tan x < cos x

(2) 2 cos 2x > 3cos x -1

I just do not know the steps .The range of x makes it worse . (Headbang) THanks for your time .

(1) Consider:

$\displaystyle \tan x = \cos x \Rightarrow \frac{\sin x}{\cos x} = \cos x$

$\displaystyle \Rightarrow \sin x = \cos^2 x \Rightarrow \sin x = 1 - \sin^2 x$

$\displaystyle \Rightarrow \sin^2 x + \sin x - 1 = 0$

$\displaystyle \Rightarrow \sin x = \frac{-1 + \sqrt{5}}{2}$ (the negative root solution is rejected - why?)

$\displaystyle \Rightarrow x = \theta_1 = \sin^{-1} \left( \frac{-1 + \sqrt{5}}{2}\right)$ or $\displaystyle x = \theta_2 = \pi - \sin^{-1} \left( \frac{-1 + \sqrt{5}}{2}\right)$.

This breaks the interval $\displaystyle [-\pi, \pi]$ into the several subintervals. Pick a simple value in each of the subintervals and use it to test whether the inequality is true or not for that subinterval.

(2) Done in a similar way.

Note: $\displaystyle 2 \cos (2x) = 3 \cos x - 1 \Rightarrow 2 \cos^2 x - 1 = 3 \cos x - 1 \Rightarrow 2 \cos^2 x - 3 \cos x = 0$.
• Oct 2nd 2009, 05:49 AM
thereddevils
Quote:

Originally Posted by mr fantastic
(1) Consider:

$\displaystyle \tan x = \cos x \Rightarrow \frac{\sin x}{\cos x} = \cos x$

$\displaystyle \Rightarrow \sin x = \cos^2 x \Rightarrow \sin x = 1 - \sin^2 x$

$\displaystyle \Rightarrow \sin^2 x + \sin x - 1 = 0$

$\displaystyle \Rightarrow \sin x = \frac{-1 + \sqrt{5}}{2}$ (the negative root solution is rejected - why?)

$\displaystyle \Rightarrow x = \theta_1 = \sin^{-1} \left( \frac{-1 + \sqrt{5}}{2}\right)$ or $\displaystyle x = \theta_2 = \pi - \sin^{-1} \left( \frac{-1 + \sqrt{5}}{2}\right)$.

This breaks the interval $\displaystyle [-\pi, \pi]$ into the several subintervals. Pick a simple value in each of the subintervals and use it to test whether the inequality is true or not for that subinterval.

(2) Done in a similar way.

Note: $\displaystyle 2 \cos (2x) = 3 \cos x - 1 \Rightarrow 2 \cos^2 x - 1 = 3 \cos x - 1 \Rightarrow 2 \cos^2 x - 3 \cos x = 0$.

Thanks Mr F . .

For (1) , i found x1 and x2 to be 38.17 and 141.83 respectively .

Then i draw a line with 4 'points', [ -pi , 38.17 , 141.83 , pi ] and tested them

i got (-pi,38.17) as the solution but this doesn't agree with my answer .

Did i miss out any sub intervals ? Thanks .
• Oct 2nd 2009, 06:06 AM
mr fantastic
Quote:

Originally Posted by thereddevils
Thanks Mr F . .

For (1) , i found x1 and x2 to be 38.17 and 141.83 respectively . Mr F says: These answers are in degrees. You're meant to get them in radians (see my reply in one of your other threads).

Then i draw a line with 4 'points', [ -pi , 38.17 , 141.83 , pi ] and tested them

i got (-pi,38.17) as the solution but this doesn't agree with my answer .

Did i miss out any sub intervals ? Thanks .

The solution will be the following intervals:

$\displaystyle -\pi \leq x < -\frac{\pi}{2}$, $\displaystyle 0.66624 < x < \frac{\pi}{2}$ and $\displaystyle 2.47535 < x \leq \pi$

(correct to 5 decimal places).

Drawing some graphs really does help show what's going on ....
• Oct 3rd 2009, 02:41 AM
thereddevils
Quote:

Originally Posted by mr fantastic
The solution will be the following intervals:

$\displaystyle -\pi \leq x < -\frac{\pi}{2}$, $\displaystyle 0.66624 < x < \frac{\pi}{2}$ and $\displaystyle 2.47535 < x \leq \pi$

(correct to 5 decimal places).

Drawing some graphs really does help show what's going on ....

Well thanks Mr F .. be patient with me ..

I think my problem is not at solving the inequality or drawing the graph .It's at solving the equation itself to find the intersection of the graphs , especially when its range is from -pi to pi . I don seem to understand how to find the x's when it goes clockwise (-ve)

where did the pi/2 come from ?
• Oct 3rd 2009, 09:50 PM
mr fantastic
Quote:

Originally Posted by thereddevils
Well thanks Mr F .. be patient with me ..

I think my problem is not at solving the inequality or drawing the graph .It's at solving the equation itself to find the intersection of the graphs , especially when its range is from -pi to pi . I don seem to understand how to find the x's when it goes clockwise (-ve)

where did the pi/2 come from ?

Draw the graphs of y = tan x and y = cos x. Note that tan x is undefined for x = pi/2, -pi/2 etc. ....