Solve the equation $2cos^2x+3cosx-3=0$ ?

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Answer 1 · 2017-05-10T12:53:04

$x=nxx360^@+-46.75^@$ , where $n$ is an integer.

We have here a quadratic equation in $cosx$

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

Hence to get $cosx$ we can use quadratic formula, where $a=2$ , $b=3$ and $c=-3$ and as solution of quadratic formula is

$(-b+-sqrt(b^2-4ac))/(2a)$

Hence $cosx=(-3+-sqrt(3^2-4xx2xx(-3)))/4=(-3+-sqrt33)/4$

i.e. $(-3-sqrt33)/4$ and $(-3+sqrt33)/4$

As $|(-3-sqrt33)/4|=(3+5.745)/4>1$ and hence is not possible and

we can only have $cosx=(-3+5.745)/4=2.745/4=0.6852$

and $cosx=cos46.75^@$ from scientific calculator or tables.

But as $cos(-x)=cosx$ we can have $x=46.75^@$ or $x=-46.75^@$

Further, every trigonometric ratio has a cycle of $360^@$

Hence $x=nxx360^@+-46.75^@$ , where $n$ is an integer.

Solve the equation 2cos^2x+3cosx-3=02cos^2x+3cosx-3=0?