Answer:
x = 0 + 2πk
√2 / 144
Step-by-step explanation:
cos x + cos 2x + cos 3x + cos 4x + cos 5x = 5
Cosine has a maximum of 1, so the only way 5 cosine terms can add up to 5 is if each one equals 1.
cos x = 1 → x = 0 + 2πk
cos 2x = 1 → x = 0 + πk
cos 3x = 1 → x = 0 + ⅔πk
cos 4x = 1 → x = 0 + ½πk
cos 5x = 1 → x = 0 + ⅖πk
The least common multiple of 2, 1, ⅔, ½, and ⅖ is 2.
So the solution that satisfies all five is x = 0 + 2πk.
lim(x→3) [√(5 + √(2x + 3)) − 2√2] / (x² − 9)
To solve without L'Hopital's rule, multiply by the conjugate of the numerator.
[tex]\lim_{x \to 3} \frac{\sqrt{5+\sqrt{2x+3} }\ -\ 2\sqrt{2} }{x^{2}-9} \times \frac{\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2} }{\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2}}[/tex]
[tex]\lim_{x \to 3} \frac{5+\sqrt{2x+3}\ -\ 8}{(x^{2}-9)(\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2})}[/tex]
[tex]\lim_{x \to 3} \frac{\sqrt{2x+3}\ -\ 3}{(x^{2}-9)(\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2})}[/tex]
Multiply by the conjugate of the new numerator.
[tex]\lim_{x \to 3} \frac{\sqrt{2x+3}\ -\ 3}{(x^{2}-9)(\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2})} \times\frac{\sqrt{2x+3}\ +\ 3}{\sqrt{2x+3}\ +\ 3}[/tex]
[tex]\lim_{x \to 3} \frac{2x+3-9}{(x^{2}-9)(\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2})(\sqrt{2x+3}\ +\ 3)}[/tex]
[tex]\lim_{x \to 3} \frac{2x-6}{(x^{2}-9)(\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2})(\sqrt{2x+3}\ +\ 3)}[/tex]
[tex]\lim_{x \to 3} \frac{2(x-3)}{(x-3)(x+3)(\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2})(\sqrt{2x+3}\ +\ 3)}[/tex]
[tex]\lim_{x \to 3} \frac{2}{(x+3)(\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2})(\sqrt{2x+3}\ +\ 3)}[/tex]
[tex]\frac{2}{(6)(4\sqrt{2})(6)}[/tex]
[tex]\frac{\sqrt{2}}{144}[/tex]
