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Question Number 175904 by BaliramKumar last updated on 09/Sep/22

$$$$$$\mathrm{If}1+sinx+{sin}^{2}x+{sin}^{3}x+........\mathrm{\infty }=4+2\sqrt{3},$$$$x=?$$

Commented by infinityaction last updated on 09/Sep/22

$$g.p.sum$$$$\frac{1}{1-\sin x}=4+2\sqrt{3}$$$$1-\sin x=\frac{1}{4+2\sqrt{3}}=\frac{4-2\sqrt{3}}{16-12}$$$$1-\sin x=\frac{4-2\sqrt{3}}{4}=1-\frac{\sqrt{3}}{2}$$$$\sin x=\frac{\sqrt{3}}{2}$$$$x=60°$$

Answered by Rasheed.Sindhi last updated on 09/Sep/22

$$\mathrm{AnOther}\mathrm{way}...$$$$\mathrm{L}etS=1+sinx+{sin}^{2}x+{sin}^{3}x+........\mathrm{\infty }...\left(i\right)$$$$\sin x\cdot S=sinx+{sin}^{2}x+{sin}^{3}x+........\mathrm{\infty }...\left(ii\right)$$$$\left(i\right)-\left(ii\right):$$$$S(1-\sin x)=1$$$$S=\frac{1}{1-\sin x}=4+2\sqrt{3}$$$$1-\sin x=\frac{1}{4+2\sqrt{3}}$$$$\sin x=1-\frac{1}{4+2\sqrt{3}}=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\cdot \frac{4-2\sqrt{3}}{4-2\sqrt{3}}$$$$=\frac{12-6\sqrt{3}+8\sqrt{3}-12}{16-12}$$$$=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$$$$x=60°$$

Answered by Rasheed.Sindhi last updated on 09/Sep/22

$$\ll \mathrm{AnOther}\mathrm{Approach}...\gg$$$$1+sinx+{sin}^{2}x+{sin}^{3}x+........\mathrm{\infty }=4+2\sqrt{3},x=?$$$$LetS=1+sinx+{sin}^{2}x+{sin}^{3}x+........\mathrm{\infty }$$$$S=1+sinx(1+sinx+{sin}^{2}x+{sin}^{3}x+........\mathrm{\infty }$$$$S=1+sinx.S$$$$S-sinx.S=1$$$$S=\frac{1}{1-sinx}=4+2\sqrt{3}$$$$1-sinx=\frac{1}{4+2\sqrt{3}}\cdot \frac{4-2\sqrt{3}}{4-2\sqrt{3}}$$$$sinx=1-\frac{4-2\sqrt{3}}{16-12}=\frac{4-4+2\sqrt{3}}{4}$$$$sinx=\frac{\sqrt{3}}{2}$$$$x=60°$$

Commented by peter frank last updated on 09/Sep/22

$$\mathrm{thanks}$$

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\mathrm{sir}.$$

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