How do you solve $7cosx=5cosx+sqrt3$ for $0<=x<=2pi$ ?

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Answer 1 · 2016-11-17T09:46:11

$x=pi/6$ or $(11pi)/6$

As $7cosx=5cosx+sqrt3$ , transposing $5cosx$ to LHS we get

$7cosx-5cosx=sqrt3$

or $2cosx=sqrt3$

or $cosx=sqrt3/2$

In the range $0<=x<=2pi$ , we can have $cosx=sqrt3/2$

when $x=pi/6$ or $x=2pi-pi/6=(11pi)/6$

Hence, $x=pi/6$ or $(11pi)/6$

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