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Question Number 129231 by liberty last updated on 14/Jan/21
 Given  { ((a=sin x+sin 2x)),((b=cos x+cos 2x)) :}. If (a^2 +b^2 )(a^2 +b^2 −3)=cos 2x+3cos x  then x =?
$$\:\mathrm{Given}\:\begin{cases}{{a}=\mathrm{sin}\:{x}+\mathrm{sin}\:\mathrm{2}{x}}\\{{b}=\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{2}{x}}\end{cases}.\:\mathrm{If}\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{cos}\:\mathrm{2}{x}+\mathrm{3cos}\:{x} \\ $$$$\mathrm{then}\:{x}\:=? \\ $$
Answered by MJS_new last updated on 14/Jan/21
let cos x =c  ⇒  a=(2c+1)(√(1−c^2 ))  b=2c^2 +c−1  a^2 +b^2 =2c+2  ⇒  2(c+1)(2c−1)=2c^2 +3c−1  c^2 −(1/2)c−(1/2)=0  cos x =−(1/2)∨cos x =1  ⇒  x=2nπ±((2π)/3) ∨ x=2nπ
$$\mathrm{let}\:\mathrm{cos}\:{x}\:={c} \\ $$$$\Rightarrow \\ $$$${a}=\left(\mathrm{2}{c}+\mathrm{1}\right)\sqrt{\mathrm{1}−{c}^{\mathrm{2}} } \\ $$$${b}=\mathrm{2}{c}^{\mathrm{2}} +{c}−\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}{c}+\mathrm{2} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}\left({c}+\mathrm{1}\right)\left(\mathrm{2}{c}−\mathrm{1}\right)=\mathrm{2}{c}^{\mathrm{2}} +\mathrm{3}{c}−\mathrm{1} \\ $$$${c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{c}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{cos}\:{x}\:=−\frac{\mathrm{1}}{\mathrm{2}}\vee\mathrm{cos}\:{x}\:=\mathrm{1} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{2}{n}\pi\pm\frac{\mathrm{2}\pi}{\mathrm{3}}\:\vee\:{x}=\mathrm{2}{n}\pi \\ $$
Commented by bemath last updated on 14/Jan/21
great...
$$\mathrm{great}… \\ $$

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