Solve:

[MorganNorthmore]

2cos^4+1=3cos^2x for 0 greater or equal to x and less than or equal to 2pie

 

[MarkFL]

Hello and welcome to FMH!

Let's arrange the equation as:

[MATH]2\cos^4(x)-3\cos^2(x)+1=0[/MATH]
Factor:

[MATH](2\cos^2(x)-1)(\cos^2(x)-1)=0[/MATH]
Now, equate each factor in turn to zero, and solve for \(x\)...can you proceed?

 

[topsquark]

2cos^4+1=3cos^2x for 0 greater or equal to x and less than or equal to 2pie

Pet peeve warning!! Danger Will Robinson! Danger!

It's "pi" not pie. ????

Pet Peeve over.

-Dan

 

[Dr.Peterson]

2cos^4+1=3cos^2x for 0 greater or equal to x and less than or equal to 2pie

Also, what you wrote (ignoring the spelling issue) is [MATH]2\cos^4(nothing) + 1 = 3 \cos^2(x)[/MATH] for [MATH]0\ge x[/MATH] and [MATH]\le 2\pi[/MATH].

In the equation, you forgot the argument of the first cosine; that's presumably just a slip.

The condition should be [MATH]0\le x\le 2\pi[/MATH]; what you wrote means [MATH]x\le 0 \text{ and } 0\le 2\pi[/MATH].

Be careful how you read or write math symbols!

But MarkFL has given good advice. Follow through, and show us your work.

 

[Mr. Bland]

It's "pi" not pie. ????

Pretty sure the correct spelling is "π". You wouldn't go off on someone misspelling "x" as "ecks" would you?

Tangent over.

 

[firemath]

Tangent over.

Circle under.
Secant second.
Sine mine.