Answer:
Please check the step-by-step explanation for the answer.
Step-by-step explanation:
Considering the trigonometric expression
[tex]6sin^2\left(3x\right)=-3[/tex]
Steps
[tex]\mathrm{Let:\:}\sin \left(3x\right)=u[/tex]
[tex]6u^2=-3[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}6[/tex]
[tex]\frac{6u^2}{6}=\frac{-3}{6}[/tex]
[tex]u^2=-\frac{1}{2}[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]u=\sqrt{-\frac{1}{2}},\:u=-\sqrt{-\frac{1}{2}}[/tex]
[tex]\mathrm{Simplify}\:\sqrt{-\frac{1}{2}}[/tex]
[tex]\mathrm{Apply\:radical\:rule}:\quad \sqrt{-a}=\sqrt{-1}\sqrt{a}[/tex]
[tex]=\sqrt{-1}\sqrt{\frac{1}{2}}[/tex]
[tex]\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i[/tex]
[tex]=i\sqrt{\frac{1}{2}}[/tex]
Similarly,
[tex]-\sqrt{-\frac{1}{2}}=\quad -i\sqrt{\frac{1}{2}}[/tex]
So,
[tex]u=i\sqrt{\frac{1}{2}},\:u=-i\sqrt{\frac{1}{2}}[/tex]
[tex]\mathrm{Substitute\:back}\:u=\sin \left(3x\right)[/tex]
[tex]\sin \left(3x\right)=i\sqrt{\frac{1}{2}},\:\sin \left(3x\right)=-i\sqrt{\frac{1}{2}}[/tex]
So,
[tex]\sin \left(3x\right)=i\sqrt{\frac{1}{2}}\quad :\quad \mathrm{No\:Solutions}[/tex]
and
[tex]\sin \left(3x\right)=-i\sqrt{\frac{1}{2}}\quad :\quad \mathrm{No\:Solutions}[/tex]
[tex]\mathrm{Combine\:all\:the\:solutions}[/tex]
[tex]\mathrm{No\:Solution\:for}\:x\in \mathbb{R}[/tex]
Keywords: trigonometric equation, solution
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