# Find all solutions in the interval..........

• July 16th 2008, 06:14 AM
sweetG
Find all solutions in the interval..........
Find all solutions in the interval -180o ≤ x ≤ 180o

to the equation cosx + (sin^2)x = 5/4
• July 16th 2008, 07:19 AM
Moo
Quote:

Originally Posted by sweetG
Find all solutions in the interval -180o ≤ x ≤ 180o

to the equation cosx + (sin^2)x = 5/4

Hello,

The writing $\sin^2 x$ means $(\sin(x))^2$ ;)

Remember the identity : $\cos^2(x)+\sin^2(x)=1 \implies \sin^2(x)=1-\cos^2(x)$

The equation is now :

$\cos x+1-\cos^2 x=\tfrac 54$

$\cos^2 x-\cos x+\tfrac 14=0$

$(\cos(x)-\tfrac 12)^2=0$

Solve the final :p
• July 18th 2008, 07:22 AM
sweetG
Does that mean x will equal 60 and - 60 ?

Is that right and are there any more solutions ?

I've forgotten how to do these types of questions >_<
• July 18th 2008, 10:40 AM
Simplicity
Quote:

Originally Posted by sweetG
Does that mean x will equal 60 and - 60 ?

Is that right and are there any more solutions ?

I've forgotten how to do these types of questions >_<

Yes, you are correct for the range provided.

The $\cos \theta$ function has a period of $360^o$. If no range was provided, then there would be infinite solution, some of which could be:

$\theta = 300, 420, 660, 780...$
• July 18th 2008, 10:40 AM
Moo
Quote:

Originally Posted by sweetG
Does that mean x will equal 60 and - 60 ?

Is that right and are there any more solutions ?

I've forgotten how to do these types of questions >_<

That's good ;)

From the final equation, you have that cos(x)-1/2=0, that is to say cos(x)=1/2

According to the unit circle there are only 2 values of x that satisfy this, between -180 and 180 (complete turn). It's ok
• July 19th 2008, 07:26 AM
sweetG
Thanks Moo and Air :)
It makes a lot more sense now