How do you solve $2 cos x + 1 = 0$ $2cosx+1=0$ algebraically?

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Answer 1 · 2018-06-14T15:27:27

$\frac{2 π}{3}, \frac{4 π}{3}$ $(2pi)/3, (4pi)/3$ on $0 \leq x \leq 2 π$ $0<=x<=2pi$

$2 cos x + 1 = 0$ $2cosx+1=0$

$2 cos x = - 1$ $2cosx=-1$

$cos x = - \frac{1}{2}$ $cosx=-1/2$

${cos}^{- 1} (- \frac{1}{2}) = x$ $cos^-1(-1/2) = x$

Use the unit circle to solve:

${cos}^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}, \frac{4 π}{3}$ $cos^-1(-1/2) = (2pi)/3, (4pi)/3$

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Answer 2 · 2018-06-14T15:40:29

Given:

$2 cos (x) + 1 = 0$ $2cos(x)+1=0$

Subtract 1 from both sides of the equation:

$2 cos (x) + 1 - 1 = 0 - 1$ $2cos(x)+1color(red)(-1)=0color(red)(-1)$

$2 cos (x) = - 1$ $2cos(x)=-1$

Divide both sides of the equation by 2:

$\frac{2}{2} cos (x) = - \frac{1}{2}$ $2/color(red)(2)cos(x)=-1/color(red)(2)$

$cos (x) = - \frac{1}{2}$ $cos(x)=-1/2$

Use the inverse cosine function on both sides:

${cos}^{- 1} (cos (x)) = {cos}^{- 1} (- \frac{1}{2})$ $color(red)(cos^-1)(cos(x))=color(red)(cos^-1)(-1/2)$

$x = {cos}^{- 1} (- \frac{1}{2})$ $x=cos^-1(-1/2)$

Within the domain $0 \leq x < 2 π$ $0<=x<2pi$ the above equation is true for 2 values of x:

$x = \frac{2}{3} π and x = \frac{4}{3} π$ $x = 2/3pi and x = 4/3pi$

To represent all of the possible solutions, we add integer multiples of $2 π$ $2pi$ to both answers:

$x = 2/3pi+ 2npi and x = 4/3pi + 2npi, n in ZZ$

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