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2 years ago

15

Giải phương trình : 8cos 4 x – 4cos2x + sin4x – 4 = 0

1 answer:

alex41 [277]2 years ago

5 0

Answer:

Step-by-step explanation:

$8(1+cos2x/2)^{2} -4c0s2x+sin4x-4=0\\2(1+2cos2x+cos^{2}2x )-4cos2x+sin4x-4=0\\2cos^{2} 2x+sin4x-2=0\\1+cos4x+sin4x-2=0\\cos4x+sin4x=1\\sin(4x+\pi /4)=sin\frac{\pi }{4} \\\left \{ {{4x+\frac{\pi }{4} =\frac{\pi }{4}+k2\pi } \atop {4x+\frac{\pi }{4} =\frac{3\pi }{4} +k2\pi \\$

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Answer:

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MArishka [77]

Answer:22/

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Step-by-step explanation:

Add: -19 + 9 = -10

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2 years ago

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maxonik [38]

Answer:

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2 years ago

Which of the following functions have a vertical asymptote for values of Ø such that cos Ø = 1?Select all that apply.

igomit [66]

Answer:

$y=\cot \theta$ and $y=\csc \theta$

Step-by-step explanation:

Note that if $\cos \theta=1,$ then $\sin \theta=0.$

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Vertical asymptotes have functions $y=\tan \theta,\ \ y=\cot \theta,\ \ y=\sec \theta,\ \ y=\csc \theta.$

Functions $y=\tan \theta$ and $y=\sec \theta$ have the same vertical asymptotes (when $\cos \theta =0$ ).

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3 years ago

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ipn [44]

Answer:

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And the face and denominator of the fraction must be positive

Step-by-step explanation:

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2 years ago