At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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]
[tex]cosx=\frac{1}{2}\to x=\frac{\pi}{3}+2k\pi\ \vee\ x=-\frac{\pi}{3}+2k\pi\ \ \ (k\in\mathbb{Z})[/tex]
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 :)

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.