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Number of  solution/solutions for \frac{\sin^2 4x+4 \sin^4 2x- 4 \sin^2 2x \cos^2 2x}{4-\sin^2 4x -4 sin^2 2x}=\frac{1}{9} will be in x\epsilon [0,\pi] ?

Option: 1 1

Option: 2 2

Option: 3 3

Option: 4 4

Answers (1)

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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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\frac{\sin^2 4x+4 \sin^4 2x- 4 \sin^2 2x \cos^2 2x}{4-\sin^2 4x -4 sin^2 2x}=\frac{1}{9}\\ \frac{4\sin^2 2x \cos^2 2x+4 \sin^4 2x- 4 \sin^2 2x \cos^2 2x}{4-4 \sin^2 2x \cos ^2 2x-4 sin^2 2x}=\frac{1}{9}\\ \frac{4 \sin^4 2x}{4 \cos ^2 2x -4 \sin^2 2x \cos ^2 2x}=\frac{1}{9}\\ \frac{4 \sin^4 2x}{4 \cos ^2 2x \cos ^2 2x}=\frac{1}{9}\\ \frac{ \sin^4 2x}{ \cos ^4 2x }=\frac{1}{9}\\ \tan^2 2x = (\tan \frac{\pi}{6})^2\\ 2x=n\pi \pm \frac{\pi}{6}\\ x=n \cdot \frac{\pi}{2} \pm \frac{\pi}{12}\\ x=\{ \frac{\pi}{12}, \frac{5\pi}{12},\frac{7\pi}{12},\frac{11\pi}{12} \}

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Gautam harsolia

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