If are two different valus of X lying between 0 and which satisfy…

Question

If are two different valus of X lying between 0 and which satisfy the equation $6\cos\text{x}+8\sin\text{x}=9$find the value of $\sin(\alpha+\beta).$

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Answer

We have, $6\cos\text{x}+8\sin\text{x}=9\ ...(1)$ $\Rightarrow8\sin\text{x}=9-6\cos\text{x}$ $\Rightarrow\big(8\sin\text{x}\big)^2=\big(9-6\cos\text{x}\big)^2$ $\Rightarrow64\sin^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$ $\Rightarrow64\sin^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$ $\Rightarrow64(1-\cos^2\text{x})=81+36\cos^2\text{x}-108\cos\text{x}$ $\Rightarrow64-64\cos^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$ $\Rightarrow36\cos^2\text{x}+64\cos^2\text{x}-108\cos\text{x}-108\cos\text{x}$ $\Rightarrow100\cos^2\text{x}-108\cos\text{x}+17=0\ ...(2)$ Since $\alpha,\beta$ are roots of equation ...(i) $$$\therefore\cos\alpha$ and $\cos\beta$ roots of equation ....(ii) $\because\cos\alpha+\cos\beta=\frac{17}{100}\ ...(3)$ $\text{Again},6\cos\text{x}+8\sin\text{x}=9$ $\Rightarrow6\cos\text{x}=9-8\sin\text{x}$ $\Rightarrow\big(8\cos\text{x}\big)^2=\big(9-6\sin\text{x}\big)^2$ $\Rightarrow36\cos^2\text{x}=81+64\cos^2\text{x}-144\cos\text{x}$ $\Rightarrow36(1-\sin^2\text{x})=81+64\sin^2\text{x}-144\sin\text{x}$ $\Rightarrow36-36\sin^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$ $[\because$ squaring both sides$]$ $\Rightarrow64\sin^2\text{x}+36\sin^2\text{x}-144\sin\text{x}+81-36=0$ $\Rightarrow100\sin^2\text{x}-144\sin\text{x}+45=0\ ...(4)$ $\because\sin\alpha\times\sin\beta=\frac{45}{100}\ ...(5)$ Now, $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ $=\frac{17}{100}-\frac{45}{100}$$=-\frac{28}{100}$$=-\frac{7}{25}$ [Using equation (3) and (5)] Now, $\sin(\alpha+\beta)=\sqrt{1-(\cos\text{x})^2}$ $=\sqrt{1-\Big(-\frac{7}{25}\Big)^2}$$=\sqrt{1-\frac{49}{625}}$ $=\sqrt{\frac{625-49}{625}}$ $=\sqrt{\frac{576}{625}}$$=\frac{24}{25}$ $\therefore\sin(\alpha+\beta)=\frac{24}{25}$