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iragen [17]
3 years ago
5

Find all solutions to the equation in the interval [0, 2TT). Cos x= sin 2x (1 point)

Mathematics
1 answer:
igomit [66]3 years ago
6 0

You can use the formula for the double angle of the sine to write

\sin(2x)=2\sin(x)\cos(x)

To turn the equation into

2\sin(x)\cos(x)=\cos(x)

Move everything to the left hand side to have

2\sin(x)\cos(x)-\cos(x)=0

Factor \cos(x) to have

\cos(x)(2\sin(x)-1)=0

A factor equals zero if and only if one of its factor equals zero. So, either

\cos(x)=0

or

2\sin(x)-1=0 \iff 2\sin(x)=1\iff \sin(x)=\dfrac{1}{2}

I assume that you have a table to lookup for these known values. If you have troubles solving for x, hit me up in the comments

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3 years ago
What are the exact solutions of x2 − 5x − 7 = 0, where x equals negative b plus or minus the square root of b squared minus 4 ti
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Answer:

x = \frac{5 + \sqrt{53}}{2}  or  x = \frac{5 - \sqrt{53}}{2}

Step-by-step explanation:

Given

x^2 - 5x - 7 = 0

Required

Solve for x using:

x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}

First, we need to identify a, b and c

The general form of a quadratic equation is:

ax^2 + bx + c = 0

So, by comparison with x^2 - 5x - 7 = 0

a = 1     b = -5      c = -7

Substitute these values of a, b and c in

x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}

x = \frac{-(-5) \± \sqrt{(-5)^2 - 4 * 1 * -7}}{2 * 1}

x = \frac{5 \± \sqrt{25 +28}}{2}

x = \frac{5 \± \sqrt{53}}{2}

Split the expression to two

x = \frac{5 + \sqrt{53}}{2}  or  x = \frac{5 - \sqrt{53}}{2}

To solve further in decimal form, we have

x = \frac{5 + 7.28}{2}  or  x = \frac{5 - 7.28}{2}

x = \frac{12.28}{2}  or  x = \frac{-2.28}{2}

x = 6.14 or x = -1.14

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