সমাধান কর: $\sin θ + \sin 2 θ + \sin 3 θ = 1 + \cos θ + \cos 2 θ, o < θ < π$

Created: 2 years ago | Updated: 4 months ago

Updated: 4 months ago

θ = \frac{π}{2}, \frac{π}{6}, \frac{π}{3}, \frac{π}{6}

ব্যাখ্যা

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Best Ans:

$\sin θ + s i n 2 θ + s i n 3 θ = 1 + c o s θ + c o s 2 θ$

$\Rightarrow 2 s i n 2 θ c o s θ + s i n 2 θ = 2 c o s^{2} θ + c o s θ$

$\Rightarrow s i n 2 θ (2 c o s θ + 1) = c o s θ (1 + 2 c o s θ)$

$\Rightarrow (1 + 2 c o s θ) (s i n 2 θ - c o s θ) = 0$

1 year ago

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Almumen

4 months ago

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MD.SOLAIMAN HOSAN

বিপরীত ত্রিকোণমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ (Inverse Trigonometric Functions and Trigonometric Equations)

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