# Thread: Solve Equations 0< or = x < or = 2pi

1. ## Solve Equations 0< or = x < or = 2pi

How would you solve these

Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.

Answear is in form of this
pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?

2. Originally Posted by j0nath0n3
How would you solve these

Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.

Answear is in form of this
pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?
Note that $\displaystyle \sin^2 x=1-\cos^2 x$

So we can rewrite the equation as $\displaystyle \cos^2 x -\left[1-\cos^2 x\right]=-\cos x$

This simplifies to $\displaystyle 2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0$

Now, if we make the substitution $\displaystyle z=\cos x$, the equation now becomes the quadratic equation $\displaystyle 2z^2+z-1=0\implies (2z-1)(z+1)=0$

This tells us that $\displaystyle z=-1$ or $\displaystyle z=\tfrac{1}{2}$

But $\displaystyle z=\cos x$, so now that means either $\displaystyle \cos x=\tfrac{1}{2}$ or $\displaystyle \cos x=-1$

Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?

3. Originally Posted by Chris L T521
Note that $\displaystyle \sin^2 x=1-\cos^2 x$

So we can rewrite the equation as $\displaystyle \cos^2 x -\left[1-\cos^2 x\right]=-\cos x$

This simplifies to $\displaystyle 2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0$

Now, if we make the substitution $\displaystyle z=\cos x$, the equation now becomes the quadratic equation $\displaystyle 2z^2+z-1=0\implies (2z-1)(z+1)=0$

This tells us that $\displaystyle z=-1$ or $\displaystyle z=\tfrac{1}{2}$

But $\displaystyle z=\cos x$, so now that means either $\displaystyle \cos x=\tfrac{1}{2}$ or $\displaystyle \cos x=-1$

Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?
Thats really good and I understood however my answear choices are in radian... and I dont understand why..

4. Originally Posted by j0nath0n3
Thats really good and I understood however my answear choices are in radian... and I dont understand why..
In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $\displaystyle 360\text{ degrees}=2\pi\text{ radians}$.

5. ## Re :

Originally Posted by Chris L T521
In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $\displaystyle 360\text{ degrees}=\pi\text{ radians}$.
Chris , is it a typo , i thought $\displaystyle 360\text{ degrees}=2\pi\text{ radians}$

6. Originally Posted by mathaddict
Chris , is it a typo , i thought $\displaystyle 360\text{ degrees}=2\pi\text{ radians}$
Yes, my mistake. Thank you for catching that.