How do I solve $cos2x+3cosx=-2$ ?

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Answer 1 · 2017-02-28T20:48:54

$x_1=npi, n=2k+-1, k in ZZ$

$x_2=n/3pi,mod(n,2)=0$

$cos2x+3cosx=-2$

Use the double angle formula for cosine to expand $cos2x$ and rewrite the equation in standard form

$2cos^2x-1+3cosx+2=0$

$2cos^2x+3cosx+1=0$

Let $cosx=y$

$y=(-3+-sqrt(9-4(2)(1)))/(2(2))=(-3+-1)/4$

$y=-1$ or $y=-1/2$

$cosx=-1$

$x=npi, n=2k+-1, k in ZZ$

$cosx=-1/2$

$x_2=n/3pi,mod(n,2)=0$

How do I solve cos2x+3cosx=-2cos2x+3cosx=-2?