How do you solve $cos (2 x) + cos (x) = 2$ $cos(2x)+cos(x)=2$ between 0 and 2pi?

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How do you solve $cos (2 x) + cos (x) = 2$ $cos(2x)+cos(x)=2$ between 0 and 2pi?

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Answer 1 · 2016-01-31T17:14:53

$x = 0, 2 π$ $x = 0 ,2pi$

Explanation:

To solve this type of equation rewrite in terms of cosx

now $cos 2 x = {cos}^{2} x - {sin}^{2} x$ $cos2x = cos^2x - sin^2x$

and ${sin}^{2} x = 1 - {cos}^{2} x \to cos 2 x = 2 {cos}^{2} x - 1$ $sin^2x = 1- cos^2x → cos2x = 2cos^2x - 1$

equation is now : $2 {cos}^{2} x - 1 + cos x = 2$ $2cos^2x - 1 + cosx = 2$

( collect terms to left side and equate to zero )

$2 {cos}^{2} x + cos x - 3 = 0$ $2 cos^2x + cosx -3 = 0$

[To factor : require 2 factors which multiply to -6

and sum to +1 (middle term)] these are -2 and 3 .

hence # (2cosx + 3 )(cosx - 1 ) = 0

so 2cosx + 3 = 0 or cosx - 1 =0

$2 cos x + 3 = 0 \to cos x = - \frac{3}{2} has no solution$ $2cosx + 3 = 0 → cosx = -3/2 color(black)( " has no solution ")$

$cos x - 1 = 0 \to cos x = 1 \to x = 0, 2 π$ $cosx - 1 = 0 → cosx = 1 → x = 0 , 2pi$

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How do you solve $cos (2 x) + cos (x) = 2$ $cos(2x)+cos(x)=2$ between 0 and 2pi?

1 Answer

Explanation:

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How do you solve cos(2x)+cos(x)=2cos(2x)+cos(x)=2cos(2x)+cos(x)=2 between 0 and 2pi?

1 Answer

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How do you solve $cos (2 x) + cos (x) = 2$ $cos(2x)+cos(x)=2$ between 0 and 2pi?