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Question Number 44128 by gunawan last updated on 22/Sep/18

The number of all possible 5−tuples (a_1 , a_2 , a_3 , a_4 , a_5 ) such that a_1 +a_2 sin x+a_3 cos x+a_4 sin 2x+a_5 cos 2x=0 holds for all x is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{5}−\mathrm{tuples} \\ $$$$\left({a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,\:{a}_{\mathrm{4}} ,\:{a}_{\mathrm{5}} \right)\:\mathrm{such}\:\mathrm{that}\: \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} \mathrm{sin}\:{x}+{a}_{\mathrm{3}} \mathrm{cos}\:{x}+{a}_{\mathrm{4}} \mathrm{sin}\:\mathrm{2}{x}+{a}_{\mathrm{5}} \mathrm{cos}\:\mathrm{2}{x}=\mathrm{0} \\ $$$$\mathrm{holds}\:\mathrm{for}\:\mathrm{all}\:\:\:{x}\:\:\mathrm{is} \\ $$

Answered by MrW3 last updated on 22/Sep/18

Way 1 to solve: x=0: a_1 +a_3 +a_5 =0 ...(1) x=π: a_1 −a_3 +a_5 =0 ...(2) (1)−(2)⇒a_3 =0 x=(π/2): a_1 +a_2 −a_5 =0 ...(3) x=−(π/2): a_1 −a_2 −a_5 =0 ...(4) (3)−(4)⇒a_2 =0 x=(π/4): a_1 +a_4 =0 ...(5) x=−(π/4): a_1 −a_4 =0 ...(6) (5)−(6)⇒a_4 =0 (5)+(6)⇒a_1 =0 (1)⇒a_5 =0 ⇒(a_1 ,...a_5 )=(0,0,0,0,0) Way 2 to solve: a_1 +(√(a_2 ^2 +a_3 ^2 )) sin (x+θ_1 )+(√(a_4 ^2 +a_5 ^2 )) sin (2x+θ_2 )=0 (√(a_2 ^2 +a_3 ^2 ))=0⇒a_2 =0,a_3 =0 (√(a_4 ^2 +a_5 ^2 ))=0⇒a_4 =0,a_5 =0 ⇒a_1 =0

$${Way}\:\mathrm{1}\:{to}\:{solve}: \\ $$$${x}=\mathrm{0}: \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{3}} +{a}_{\mathrm{5}} =\mathrm{0}\:\:\:...\left(\mathrm{1}\right) \\ $$$${x}=\pi: \\ $$$${a}_{\mathrm{1}} −{a}_{\mathrm{3}} +{a}_{\mathrm{5}} =\mathrm{0}\:\:\:...\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\Rightarrow{a}_{\mathrm{3}} =\mathrm{0} \\ $$$$ \\ $$$${x}=\frac{\pi}{\mathrm{2}}: \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} −{a}_{\mathrm{5}} =\mathrm{0}\:\:\:...\left(\mathrm{3}\right) \\ $$$${x}=−\frac{\pi}{\mathrm{2}}: \\ $$$${a}_{\mathrm{1}} −{a}_{\mathrm{2}} −{a}_{\mathrm{5}} =\mathrm{0}\:\:\:...\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right)−\left(\mathrm{4}\right)\Rightarrow{a}_{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$$${x}=\frac{\pi}{\mathrm{4}}: \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{4}} =\mathrm{0}\:\:\:...\left(\mathrm{5}\right) \\ $$$${x}=−\frac{\pi}{\mathrm{4}}: \\ $$$${a}_{\mathrm{1}} −{a}_{\mathrm{4}} =\mathrm{0}\:\:\:...\left(\mathrm{6}\right) \\ $$$$\left(\mathrm{5}\right)−\left(\mathrm{6}\right)\Rightarrow{a}_{\mathrm{4}} =\mathrm{0} \\ $$$$\left(\mathrm{5}\right)+\left(\mathrm{6}\right)\Rightarrow{a}_{\mathrm{1}} =\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\Rightarrow{a}_{\mathrm{5}} =\mathrm{0} \\ $$$$\Rightarrow\left({a}_{\mathrm{1}} ,...{a}_{\mathrm{5}} \right)=\left(\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0}\right) \\ $$$$ \\ $$$${Way}\:\mathrm{2}\:{to}\:{solve}: \\ $$$${a}_{\mathrm{1}} +\sqrt{{a}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} }\:\mathrm{sin}\:\left({x}+\theta_{\mathrm{1}} \right)+\sqrt{{a}_{\mathrm{4}} ^{\mathrm{2}} +{a}_{\mathrm{5}} ^{\mathrm{2}} }\:\mathrm{sin}\:\left(\mathrm{2}{x}+\theta_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\sqrt{{a}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} }=\mathrm{0}\Rightarrow{a}_{\mathrm{2}} =\mathrm{0},{a}_{\mathrm{3}} =\mathrm{0} \\ $$$$\sqrt{{a}_{\mathrm{4}} ^{\mathrm{2}} +{a}_{\mathrm{5}} ^{\mathrm{2}} }=\mathrm{0}\Rightarrow{a}_{\mathrm{4}} =\mathrm{0},{a}_{\mathrm{5}} =\mathrm{0} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\mathrm{0} \\ $$

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