Find the solutions of the equation that are in the interval [0,2pi/3)? cos3u+cos6u=0

Question

#cos3u+cos6u=0#

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Answer 1 · 2018-04-12T11:56:59

#u=pi/9,pi/3,(5pi)/9. or u=20^circ,60^circ,100^circ.#
Hint: #color(red)(cosC+cosD=2cos((C+D)/2)cos((C-D)/2))#

Here,

#cos3u+cos6u=0#

#=>2cos((3u+6u)/2)cos((3u-6u)/2)=0#

#=>cos((9u)/2)cos((-3u)/2)=0#

#=>cos((9u)/2)=0 or cos((3u)/2)=0#

We have two options;

#(I)cos((9u)/2)=0=>(9u)/2=pi/2,(3pi)/2,(5pi)/2,(7pi)/2,(9pi)/2...#

#=>9u=pi,3pi,5pi,7pi,9pi,...#

#=>u=pi/9,(3pi)/9,(5pi)/9,(7pi)/9,(9pi)/9,...#

But, #uin [0,(2pi)/3)#i.e. #uin[0^circ,120^circ)#

#:.u=pi/9,(3pi)/9,(5pi)/9, are # possible.

#:.u=pi/9,pi/3,(5pi)/9.#

#(II)cos((3u)/2)=0=>(3u)/2=pi/2,(3pi)/2,(5pi)/2,(7pi)/2,(9pi)/2...#

#=>3u=pi,3pi,5pi,7pi,9pi,...#

#=>u=pi/3,(3pi)/3,(5pi)/3,...#

But #uin [0,(2pi)/3)=>u=pi/3 #, is possible.

From #(I) and(II)# we have #u=pi/9,pi/3,(5pi)/9.#