Find all solutions of the equation 2sin2x−cosx=1 in the interval [0,2π), what is x?

1 answer:

4 0

This is a little it hard... I am wondering if others have a better way to solve it.

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2sin2x−cosx=1

4sinxcosx = 1 + cosx

16sin^2xcos^2x = 1 + 2cosx + cos^2x

16(1 - cos^2x)cos^2x = 1 + 2cosx + cos^2x

16cos^2x - 16cos^4x = 1 + 2cosx + cos^2x

16cos^4x - 15cos^2x + 2cosx + 1 = 0

let cosx = Y

16Y^4 - 15Y^2 + 2Y + 1 = 0

(Y + 1)(16Y^3 - 16Y^2 + Y + 1) = 0

there are four values for Y, -1, -0.2, 0.367, 0.836.

Then you solve for Y.

For example, Y = -1 means cosx = -1, x = π.

<span>I will leave the rests to you. (Sorry, this seems to be an ugly way...)</span>

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3 years ago

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steposvetlana [31]

Answer:

on rearranging,

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3 years ago

Rationalize the denominator 3-sqrt(6)÷4+sqrt(5)

Ivan

Answer:

$\dfrac{12-3\sqrt{5}-4\sqrt{6}+\sqrt{30}}{11}$

Step-by-step explanation:

You have written 3 - ((√6)/4) + √5. The only denominator is 4, which is already rational. If you were to use appropriate grouping symbols (parentheses) we might expect to see ...

(3 -√6)/(4 +√5)

The denominator is rationalized by taking advantage of the factoring of the difference of squares:

a² - b² = (a -b)(a +b)

The denominator will be rational if we can square the value √5. By choosing a multiplier of (4-√5), we can do that:

$\dfrac{3-\sqrt{6}}{4+\sqrt{5}}=\dfrac{(3-\sqrt{6})(4-\sqrt{5})}{(4+\sqrt{5})(4-\sqrt{5})}\\\\=\dfrac{3\cdot 4-3\sqrt{5}-4\sqrt{6}+\sqrt{30}}{4^2-(\sqrt{5})^2}=\dfrac{12-3\sqrt{5}-4\sqrt{6}+\sqrt{30}}{11}$