Solve For X: Sin X + Sin 2x + Sin 3x = Cos X + Cos 2x + Cos 3x In ... - Toppr

SolveGuidesJoin / LoginUse appLogin0You visited us 0 times! Enjoying our articles? Unlock Full Access!Standard XIIMathematicsQuestionSolve for x:sinx+sin2x+sin3x=cosx+cos2x+cos3x in the interval 0x2π.Open in AppSolutionVerified by Toppr

We write the given equation as(sinx+sin3x)+sin2x=(cosx+cos3x)+cos2xor 2sin2xcosx+sin2x=2cos2xcosx+cos2xor sin2x(2cosx+1)=cos2x(2cosx+1)or (sin2xcos2x)(2cosx+1)=0sin2xcos2x=0 or 2cosx+1=0If sin2xcos2x=0, then tan2x=1,Hence 2x=nπ+π/4or x=(4n+1)π8 .(1)If 2cosx+1=0, then cosx=1/2 (2)x=2nπ±2π3 or x=6n±23πWe seek values of x in the interval 0x2πIn this interval (1) givesx=π/8,5π/8,9π/8,13π/8. (n=0,1,2,3)and (2) gives x=2π/3,4π/3. (for n=0,1)Thus we get the answerx=π/8,5π/8,2π/3,9π/8,4π/3,13π/8.

Was this answer helpful?18Similar QuestionsQ1Solve for x:sinx+sin2x+sin3x=cosx+cos2x+cos3x in the interval 0x2π.View SolutionQ2Solve the following equations: (i) cos x+cos 2x+cos 3x=0 (ii) cos x+cos 3x-cos 2x=0 (iii) sin x+sin 5x=sin 3x (iv) cos x cos 2x cos 3x=14 (v) cos x+sin x=cos 2x+sin 2x (vi) sin x+sin 2x+sin 3=0 (vii) sin x+sin 2x+sin 3x+sin 4x=0 (viii) sin 3x-sin x=4 cos2 x-2 (ix) sin 2x-sin 4x+sin 6x=0View SolutionQ3Solve cos3x+cos2x=sin(3x/2)+sin(x/2),0x<2π.View SolutionQ4

The number of solution of the equation sinx+sin2x+sin3x=cosx+cos2x+cos3x,0x2π is

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Q5Solve the following equations.sinx+sin2x+sin3x=cosx+cos2x+cos3x.View Solution

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