If 2 sin 2-3 cos- where 0 - 2 - then find the value of -

2sin^2θ=3 cosθ 2(1 cos^2θ)=3 cosθ [∵ sin^2θ=1 cos^2θ] 2 2 cos^2θ=3 cos θ ⇒ 2 cos^2θ+3 cos θ 2=0 ⇒ 2 cos^2θ+4 cos θ cosθ 2=0 ⇒ 2 cos θ[cosθ+2] [cosθ+ ...

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Standard XII

Mathematics

Equations with Solutions at Boundary Values

If 2 sin 2θ=3...

Question

$I f 2 s i n^{2} θ = 3 c o s θ, w h e r e 0 \leq θ \leq 2 π, t h e n f i n d t h e v a l u e o f θ .$

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Solution

$2 s i n^{2} θ = 3 c o s θ 2 (1 - c o s^{2} θ) = 3 c o s θ [∵ s i n^{2} θ = 1 - c o s^{2} θ] 2 - 2 c o s^{2} θ = 3 c o s θ \Rightarrow 2 c o s^{2} θ + 3 c o s θ - 2 = 0 \Rightarrow 2 c o s^{2} θ + 4 c o s θ - c o s θ - 2 = 0 \Rightarrow 2 c o s θ [c o s θ + 2] - [c o s θ + 2] = 0 \Rightarrow (2 c o s θ - 1) (c o s θ + 2) = 0 \Rightarrow 2 c o s θ - 1 = 0 o r c o s θ = - 2 ∵ θ = \frac{π}{3,} \frac{5 π}{3}$

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If 2sin2θ=3 cosθ,where 0≤θ≤ 2π,then find the value of θ.

2sin2θ=3 cosθ2(1−cos2θ)=3 cosθ [∵sin2θ=1−cos2θ]2−2 cos2θ=3 cos θ⇒2 cos2θ+3 cos θ−2=0⇒2 cos2θ+4 cos θ−cosθ−2=0⇒2 cos θ[cosθ+2]−[cosθ+2]=0⇒(2 cos θ−1)(cos θ+2)=0⇒2 cos θ −1=0 or cosθ=−2∵ θ=π3,5π3

$I f 2 s i n^{2} θ = 3 c o s θ, w h e r e 0 \leq θ \leq 2 π, t h e n f i n d t h e v a l u e o f θ .$