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2cos^2(x)-5cos(x)+2=0

Sagot :

[tex]2cos^2x-5cos+2=0\\\\cosx=t\in < -1;\ 1 >\\\\2t^2-5t+2=0\\\\a=2;\ b=-5;\ c=2\\\\\Delta=b^2-4ac;\ \Delta=(-5)^2-4\cdot2\cdot2=25-16=9\\\\t_1=\frac{-b-\sqrt\Delta}{2a};\ t_2=\frac{-b+\sqrt\Delta}{2a}\\\\t_1=\frac{5-\sqrt9}{2\cdot2}=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}\in < -1;\ 1 >\\\\t_2=\frac{5+\sqrt9}{2\cdot2}=\frac{5+3}{4}=\frac{8}{4}=2\notin < -1;\ 1 >[/tex]

[tex]cosx=\frac{1}{2}\to x=\frac{\pi}{3}+2k\pi\ \vee\ x=-\frac{\pi}{3}+2k\pi\ \ \ (k\in\mathbb{Z})[/tex]
Louli

Answer:

[tex] x = \frac{\pi}{3} + 2n\pi \\ or\\ x=\frac{-\pi}{3} + 2n\pi [/tex]

where "n" is an integer that belongs to Z.

Explanation:

The equation given is:

2cos²(x) - 5cos(x) + 2 = 0

To factor this equation, we will use the quadratic formula shown in the attached image.

From the given equation:

a = 2

b = -5

c = 2

This means that:

either cos(x) = [tex] \frac{5+\sqrt{(-5)^2-4(2)(2)}}{2(2)} = 2 [/tex] .......> This solution is rejected as the value of the cosine function lies between -1 and 1 only.

or cos(x) = [tex] \frac{5-\sqrt{(-5)^2-4(2)(2)}}{2(2)} = 0.5 [/tex] ......> This solution is accepted as it lies within -1 and 1

Now, using the inverse of the cosine, we can find that:

[tex] x = \frac{\pi}{3} + 2n\pi \\ or\\ x=\frac{-\pi}{3} + 2n\pi [/tex]

where "n" is an integer that belongs to Z.

Hope this helps :)

View image Louli
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