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padilas [110]
3 years ago
9

Solve the equation 2sin(x) + 1= 0

Mathematics
1 answer:
noname [10]3 years ago
8 0

Answer:

Step-by-step explanation:

+ First, we write 2sin(x) = -1 (add -1 for both sides)

Then sin(x) = -1/2 (divide both sides by 2).

+ And we know that -1/2 = sin( -pi/6)

So sin(x) = sin (-pi/6)

that implicite:

x=-\frac{\pi }{6} + 2k\pi\\x= \pi - (-\frac{\pi}{6} )+ 2k\pi= \frac{7\pi}{6} +2k\pi

Hope it useful for you.

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Which are the solutions of the quadratic equation x^2=-5x-3
frutty [35]
The correct answer would be \frac{-5 +/-  \sqrt{13} }{2}

In order to solve for this you must first get the equation equal to 0. 

<span>x^2=-5x-3 ----> add 5x to both sides. 
x^2 + 5x = -3 ----> add 3 to both sides.
x^2 + 5x + 3 = 0

Now knowing this we can use the coefficients of each one in descending order of power as a, b and c. 

a = 1 (because it is the coefficient to x^2)
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c = 3 (because it is the end number)

Now we can plug these values into the quadratic equation. 

\frac{-b +/-  \sqrt{b^{2} - 4ac } }{2a}

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And those would be your two answers.
5 0
3 years ago
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Answer:

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