7 cos x+1=6 secx solve for x

1 answer:

6 0

7cos(x) + 1 = 6sec(x)
7cos(x) + 1 = 6/cos(x)
7cos^(x) + cos(x) = 6
7cos^(x) + cos(x) - 6 = 0
[7cos(x) - 6][cos(x) + 1] = 0
cos(x) = 6/7 , x = arccos(6/7) and
cos(x) = -1, x = 180

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Answer:

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Step-by-step explanation:

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2 years ago

The roots of 2x^2-3x=4 are a and b. Find the simplest quadratic equation which has roots 1/a and 1/b

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Find the roots

solve
we use hmm, completing the suare
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x²-1.5x=2
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-1.5/2=-0.75, (-0.75)²=0.5625
add that to both sides
x²-1.5x+0.5625=2+0.5625
factor perfect squaer trinomial
(x-0.75)²=2.5625
square root both sides, remember to take positive and negative square roots
x-0.75=+/-√2.5625
add 0.75 to both sides
x=0.75+/-√2.5625

the roots are x=0.75+√2.5625 and x=0.75-√2.5625

1/a and 1/b
1/(0.75+√2.5625) and 1/(0.75-√2.5625)

if the roots of a quadratic equation are r1 and r2 then it factors to
(x-r1)(x-r2)
so then we can factor our equation to be
$(x-\frac{1}{0.75+\sqrt{2.5625}})(x-\frac{1}{0.75-\sqrt{2.5625}})$

if we were to try and expand it, we would get
x²+0.75x-0.5
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x²+0.75x-0.5 is answer

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Vilka [71]

Answer:

The inequality describing the situation is:

$7b+9u>1500$