If cos 3 x-cos 2 x-sin3 x-2-sinx-2- 0 - x - 2 - thenx -

The correct option is A cos 3x + cos2x =sin (3x/2)+sin (x/2) ⇒ 2 cos (5x/2) cos (x/2) =2 sin x. cos (x/2) ⇒ cos x/2=0 or cos 5x/2=sin x = cos(90 x)⇒ x = π, π/7 ...

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Byju's Answer

Standard XI

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

If cos 3 x+co...

Question

If $c o s 3 x + c o s 2 x = s i n (\frac{3 x}{2}) + s i n (\frac{x}{2}), 0 \leq x \leq 2 π, t h e n x =$

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Solution

The correct option is A
$c o s 3 x + c o s 2 x = s i n (\frac{3 x}{2}) + s i n (\frac{x}{2}) \Rightarrow 2 c o s (\frac{5 x}{2}) c o s (\frac{x}{2}) = 2 s i n x . c o s (\frac{x}{2}) \Rightarrow c o s \frac{x}{2} = 0 o r c o s \frac{5 x}{2} = s i n x = c o s (90 - x) \Rightarrow x = π, \frac{π}{7}, \frac{5 π}{7}, \frac{9 π}{7}, \frac{13 π}{7}$

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If cos3x+cos2x=sin(3x2)+sin(x2),0≤x≤2π,thenx=

The correct option is A cos3x+cos2x=sin(3x2)+sin(x2)⇒2cos(5x2)cos(x2)=2sinx.cos(x2)⇒cosx2=0orcos5x2=sinx=cos(90−x)⇒x=π,π7,5π7,9π7,13π7

If $c o s 3 x + c o s 2 x = s i n (\frac{3 x}{2}) + s i n (\frac{x}{2}), 0 \leq x \leq 2 π, t h e n x =$