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  • Find the number of solutions for <img alt="\sin^{10}\theta+\cos^{10}\theta=\cos^4 2\theta \ \ \ \theta\epsilon[0,{\pi}]" src="https://learn.careers360.com/latex-image/?%5Csin%5E%7B10%7D%5Cthet

Find the number of solutions for \sin^{10}\theta+\cos^{10}\theta=\cos^4 2\theta \ \ \ \theta\epsilon[0,{\pi}]

Option: 1 0

Option: 2 1

Option: 3 2

Option: 4 4

Answers (1)

best_answer

General Solution of some Standard Equations (Part 2) -

General Solution of some Standard Equations (Part 2)

 

4. sin2 ? = sin2 α

\begin{array}{l}{\Rightarrow \quad \sin ^{2} \theta=\sin ^{2} \dot{\alpha}} \\ {\Rightarrow \quad \sin (\theta+\alpha) \sin (\theta-\alpha)=0} \\\mathrm{\because we\;are\;using\;the\;identity,\sin (A+B) \sin (A-B)=\sin ^{2} A-\sin ^{2} B}\\ {\Rightarrow \sin (\theta+\alpha)=0 \text { or } \sin (\theta-\alpha)=0} \\ {\Rightarrow \theta+\alpha=n \pi \text { or } \theta-\alpha=n \pi, n \in \mathbb{I}} \\ {\Rightarrow \theta=n \pi \pm \alpha \in \mathbb{I}}\end{array}

The general solution of the equation cos2 ? = cos2 α is  \mathrm{{ \theta=n \pi \pm \alpha \in \mathbb{I}}} .

 

And, the general solution of the equation tan2 ? = tan2 α is  \mathrm{{ \theta=n \pi \pm \alpha \in \mathbb{I}}} .

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\sin^{10}\theta+\cos^{10}\theta=\cos^4 2\theta \ \ \ \theta\epsilon[0,\pi] \\ (\frac{1-\cos 2\theta}{2})^5+(\frac{1+\cos 2\theta}{2})^5=\cos^4 2\theta\\ put\ \cos2 \theta=x\\ (\frac{1-x}{2})^5+(\frac{1+x}{2})^5= x^4\\ (a+b)^{5}=a^{5}+5 \cdot a^{4} \cdot b+10 \cdot a^{3} \cdot b^{2}+10 \cdot a^{2} \cdot b^{3}+5 \cdot a^{1} \cdot b^{4}+b^{5}\\ 1-5x+10x^2-10 x^3+5x^4-x^5+1+5x+10x^2+10x^3+5x^4+x^5=32x^4\\ 22x^4-20 x^2-2=0\\ 11x^4-10 x^2-1=0\\ (x^2-1)(11x^2+1)=0\\ only\ real\ root\ =\pm 1 or\ \ x^2=1\\ cos^2 2\theta=1\\ 2\theta=n \pi \pm \frac{\pi}{2}\\ \theta=\{ \frac{\pi}{4},\frac{3\pi}{4}\}

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