If cos 3 -cos 4 -cos 5 -cos 6 - then -

The correct option is C cos 3 θ cos 4θ = cos 5 θ cos 6 θ⇒ cos 3 θ cos 5 θ = cos 4 θ cos 6 θ ⇒ 2 sin 4 θ. sin θ = 2 sin 5θ. sin θ⇒ sin θ (sin 5 θ sin ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Equations with Solutions at Boundary Values

If cos 3 θ-co...

Question

If $c o s 3 θ - c o s 4 θ = c o s 5 θ - c o s 6 θ t h e n θ =$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$c o s 3 θ - c o s 4 θ = c o s 5 θ - c o s 6 θ \Rightarrow c o s 3 θ - c o s 5 θ = c o s 4 θ - c o s 6 θ$
$\Rightarrow 2 s i n 4 θ . s i n θ = 2 s i n 5 θ . s i n θ \Rightarrow s i n θ (s i n 5 θ - s i n 4 θ) = 0 \Rightarrow 2 s i n θ . c o s \frac{9 θ}{2} s i n \frac{θ}{2} = 0$
$\Rightarrow s i n θ = 0, c o s \frac{9 θ}{2} = 0, \Rightarrow n π o r 2 n π o r (2 n + 1) \frac{π}{9}$

Suggest Corrections

If cos3θ−cos4θ=cos5θ−cos6θthenθ=

The correct option is C cos3θ−cos4θ=cos5θ−cos6θ⇒cos3θ−cos5θ=cos4θ−cos6θ ⇒2sin4θ.sinθ=2sin5θ.sinθ⇒sinθ(sin5θ−sin4θ)=0⇒2sinθ.cos9θ2sinθ2=0 ⇒sinθ=0,cos9θ2=0,⇒nπor2nπor(2n+1)π9

If $c o s 3 θ - c o s 4 θ = c o s 5 θ - c o s 6 θ t h e n θ =$