Prove x -2 n-2 -3 orx - m -1 m -7 -6- where m - n - I

4 sin x cos x+2 sin x+2 cos x+1=0 ⇒ 2 sin x (2 cos x +1)+(2 cos x +1)=0 ⇒ (2 cos x +1)(2 sin x +1)=0 ⇒ cos x = 1/2 or sin x = 1/2 ⇒ cos x = 1/2= c ...

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Standard XI

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

Prove x =2 nπ...

Question

Prove $x = (2 n π \pm \frac{2 π}{3}) o r x = m π + (- 1)^{m} . \frac{7 π}{6}$ , where m, $n \in I$

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Solution

$4 s i n x c o s x + 2 s i n x + 2 c o s x + 1 = 0$

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

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

$\Rightarrow c o s x = - \frac{1}{2}$ or $s i n x = - \frac{1}{2}$

$\Rightarrow c o s x = - \frac{1}{2} = - c o s \frac{π}{3} = c o s (π - \frac{π}{3}) = c o s \frac{2 π}{3}$

or $s i n x = - \frac{1}{2} = - s i n \frac{π}{6} = s i n (π + \frac{π}{6}) = s i n \frac{7 π}{6}$

$\Rightarrow x = (2 n π \pm \frac{2 π}{3})$ or $x = m π + (- 1)^{m} . \frac{7 π}{6}$ , where m, $n \in I .$

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Prove x=(2nπ±2π3)orx=mπ+(−1)m.7π6, where m, n∈I

4 sin x cos x+2 sin x+2 cos x+1=0 ⇒ 2 sin x (2 cos x+1)+(2 cos x+1)=0 ⇒(2 cos x+1)(2 sin x+1)=0 ⇒cos x=−12 or sin x=−12 ⇒cos x=−12=−cos π3=cos (π−π3)=cos 2π3 or sin x=−12=−sin π6=sin (π+π6)=sin 7π6 ⇒x=(2 n π ± 2 π3) or x=m π+(−1)m.7π6, where m, n∈I.

Prove $x = (2 n π \pm \frac{2 π}{3}) o r x = m π + (- 1)^{m} . \frac{7 π}{6}$ , where m, $n \in I$