If 32 tan 8-2 cos 2-3 cos- and 3 cos 3-6 cos 2-2 cos-4-0- then general values of - is

The correct option is D 2n π±2π/3, n ϵ Z3 cos^3 θ + 6 cos^2 θ 2 cos θ 4 =0 ⇒ (3 cos^2 θ 2)(cos θ + 2)=0 ∴ cos^2 θ =2/3⇒ sec^2 θ =3/2 ∴ tan^2 θ =1/2⇒ tan^8 ...

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Byju's Answer

Standard XII

Mathematics

Property 1

If 32 tan 8θ=...

Question

If $32 t a n^{8} θ = 2 c o s^{2} α - 3 c o s α a n d 3 c o s^{3} θ + 6 c o s^{2} θ - 2 c o s θ - 4 = 0$ , then general values of $α$ is

n π \pm \frac{2 π}{3}, n ϵ Z

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n π + \frac{2 π}{3}, n ϵ Z

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2 n π \pm \frac{2 π}{3}, n ϵ Z

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2 n π + \frac{2 π}{3}, n ϵ Z

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Solution

The correct option is C $2 n π \pm \frac{2 π}{3}, n ϵ Z$
$3 c o s^{3} θ + 6 c o s^{2} θ - 2 c o s θ - 4 = 0 \Rightarrow (3 c o s^{2} θ - 2) (c o s θ + 2) = 0 ∴ c o s^{2} θ = \frac{2}{3} \Rightarrow s e c^{2} θ = \frac{3}{2} ∴ t a n^{2} θ = \frac{1}{2} \Rightarrow t a n^{8} θ = \frac{1}{16} ∴ 32 t a n^{8} θ = 2 = 2 c o s^{2} α - 3 c o s α$
$\Rightarrow c o s α = - \frac{1}{2} o r c o s α = 2 (N o t p o s s i b l e) ∴ c o s α = - \frac{1}{2} = c o s \frac{2 π}{3} ∴ α = 2 n π \pm \frac{2 π}{3}$

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If 32 tan8θ=2 cos2α−3 cos α and 3 cos3θ+6 cos2θ−2 cos θ−4=0, then general values of α is

The correct option is C 2nπ±2π3, nϵZ3 cos3θ+6 cos2θ−2 cos θ−4=0⇒ (3 cos2θ−2)(cos θ+2)=0∴ cos2θ=23⇒sec2θ=32∴ tan2θ=12⇒tan8θ=116∴ 32 tan8θ=2=2 cos2α−3 cos α ⇒ cos α=−12 or cos α=2 (Not possible)∴ cos α=−12=cos 2π3∴ α=2nπ±2π3

If $32 t a n^{8} θ = 2 c o s^{2} α - 3 c o s α a n d 3 c o s^{3} θ + 6 c o s^{2} θ - 2 c o s θ - 4 = 0$ , then general values of $α$ is