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earnstyle [38]
3 years ago
12

Find all solutions to the equation. cos2x + 2 cos x + 1 = 0

Mathematics
1 answer:
yanalaym [24]3 years ago
5 0
For cos(2x) * (2cos(x) + 1) = 0, use the double angle identity for cos(2x), which is cos^2 x - sin^2 x = cos^2 x - (1-cos^2) = 2cos^2 x - 1.
So we have (2cos^2 x - 1)(2cos x + 1) = 0. So 2cos^2 x -1 = 0 or x = 0 and 2pi.
For 2sec^2 x + tan^2 x - 3 = 0, use the identity sec^2 x = tan^2 x + 1, so we have
2(tan^2 x + 1) + tan^2 x - 3 = 0 or
<span>2tan^2 x + tan^2 x - 1 = 0 or
</span>3 tan^2 x = 1.

So x = pi/2, pi/2 + pi = 3pi/2.

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F(x)=4x+1 and g(x)=x^2-5find(f-g)(x)
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Answer:

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Given f(x)=4x+1 and g(x)=x²-5

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Step-by-step explanation:

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x^2 + 4x + y^2 +8y  =  0

Step-by-step explanation:

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Open brackets

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Collect like terms

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3x^2 + 12x + 3y^2 +24y  =  0

Divide through by 3

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