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Question Number 118506 by meireza last updated on 18/Oct/20

Answered by Lordose last updated on 18/Oct/20

$$\sqrt{2}\sin (\mathrm{x}°+15°)=1$$$$\mathrm{Divide}\mathrm{by}\sqrt{2}$$$$\sin (\mathrm{x}°+15°)=\frac{1}{\sqrt{2}}$$$$\mathrm{take}{\sin }^{-1}\mathrm{of}\mathrm{both}\mathrm{sides}$$$${\sin }^{-1}\left(\sin (\mathrm{x}+15°)\right)={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$$$$\mathrm{x}+15°=45°+360\mathrm{n}\{\mathrm{n}\in \mathbb{N}\}$$$$\mathrm{x}=30°+360\mathrm{n}$$$$\mathrm{x}=\frac{\pi }{6}+2\pi \mathrm{n}\{\mathrm{n}\in \mathbb{N}\}$$

Answered by ebi last updated on 18/Oct/20

$$$$$$$$$$\sqrt{2}\sin (x+15)°=1$$$$\sin (x+15)°=\frac{1}{\sqrt{2}}$$$$let\theta =x+15,$$$$\sin \theta °=\frac{1}{\sqrt{2}}$$$$sincethevalueispositive,thevaluesof\theta aresituatedatquadrantIandII.$$$$therefore,$$$$\theta °=45°,180°-45°,$$$$\theta °=45°,135°$$$$$$$$thus,$$$$x+15°=45°,135°$$$$x=30°,120°(0°⩽x⩽360°)$$

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