Question #235813

Find the smallest value of parameter alpha such that equation

(sin(x))^2 * cos(2x) + alpha*(cos(x)^4-sin(x)^4)= -10(2* alpha +1)^2

has at least one real solution.


1
Expert's answer
2021-09-16T00:46:57-0400
cos4xsin4x=(cos2xsin2x)(cos2x+sin2x)\cos^4x-\sin^4x=(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)

=cos(2x)1=cos(2x)=\cos(2x)\cdot1=\cos(2x)

sin2x=1cos(2x)2\sin^2x=\dfrac{1-\cos(2x)}{2}

1cos(2x)2(cos(2x))+αcos(2x)=10(2α+1)2\dfrac{1-\cos(2x)}{2}(\cos(2x))+\alpha\cos(2x)=-10(2\alpha+1)^2

cos2(2x)(1+2α)cos(2x)20(2α+1)2=0\cos^2(2x)-(1+2\alpha)\cos(2x)-20(2\alpha+1)^2=0

1cos(2x)1-1\leq \cos(2x)\leq 1

D=((1+2α))24(1)(20(2α+1)2)=81(2α+1)2D=(-(1+2\alpha))^2-4(1)(-20(2\alpha+1)^2)=81(2\alpha+1)^2

cos(2x)=(1+2α)±9(1+2α)2\cos(2x)=\dfrac{(1+2\alpha)\pm9(1+2\alpha)}{2}

cos(2x)=4(1+2α) or cos(2x)=5(1+2α)\cos(2x)=-4(1+2\alpha)\text{ or } \cos(2x)=5(1+2\alpha)

14(1+2α)1-1\leq-4(1+2\alpha)\leq1

141+2α14-\dfrac{1}{4}\leq1+2\alpha\leq\dfrac{1}{4}

542α34-\dfrac{5}{4}\leq2\alpha\leq -\dfrac{3}{4}

58α38-\dfrac{5}{8}\leq\alpha\leq -\dfrac{3}{8}


15(1+2α)1-1\leq5(1+2\alpha)\leq1

151+2α15-\dfrac{1}{5}\leq1+2\alpha\leq\dfrac{1}{5}

652α45-\dfrac{6}{5}\leq2\alpha\leq -\dfrac{4}{5}

35α25-\dfrac{3}{5}\leq\alpha\leq -\dfrac{2}{5}

58=2540,35=2440-\dfrac{5}{8}=-\dfrac{25}{40}, -\dfrac{3}{5}=-\dfrac{24}{40}



The smallest value of parameter α\alpha is 58-\dfrac{5}{8} .


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