The number of values of - in -0-2 - that satisf y the equation sin 2-cos-1-4

The correct option is B 2 sin^2θ cosθ=1/4 ⇒ (1 cos^2θ) cosθ= 1/4 ⇒ 4 4cos^2θ 4cosθ=1 ⇒ 4 cos^2θ+4cosθ 3=0 ⇒ 4 cos^2θ+6cosθ 2 cosθ 3=0 ⇒ 2cosθ(2cosθ+3) 1(2cosθ+ ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

The number of...

Question

$The number of values of θ i n [0, 2 π] t h a t s a t i s f y t h e e q u a t i o n s i n^{2} θ - c o s θ \frac{1}{4}$

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Solution

The correct option is B
2

$s i n^{2} θ - c o s θ = \frac{1}{4} \Rightarrow (1 - c o s^{2} θ) - c o s θ = \frac{1}{4} \Rightarrow 4 - 4 c o s^{2} θ - 4 c o s θ = 1 \Rightarrow 4 c o s^{2} θ + 4 c o s θ - 3 = 0 \Rightarrow 4 c o s^{2} θ + 6 c o s θ - 2 c o s θ - 3 = 0 \Rightarrow 2 c o s θ (2 c o s θ + 3) - 1 (2 c o s θ + 3) = 0 \Rightarrow (2 c o s θ + 3) (2 c o s θ - 1) = 0 \Rightarrow 2 c o s θ + 3 = 0 o r, 2 c o s θ - 1 = 0 \Rightarrow c o s θ = - \frac{3}{2} o r c o s θ = \frac{1}{2} H e r e, c o s θ = - \frac{3}{2} i s n o t p o s s i b l e . ∴ c o s θ = \frac{1}{2} \Rightarrow c o s θ = c o s \frac{π}{3} \Rightarrow θ = 2 n π \pm \frac{π}{3} N o w f o r n = 0 a n d 1, t h e v a l u e s o f θ a r e \frac{π}{3}, \frac{5 π}{3} a n d \frac{7 π}{3}, b u t \frac{7 π}{3} i s n o t i n [o, 2 π] H e n c e, t h e r e a r e t w o s o l u t i n o s i n [o, 2 π] .$

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The number of values of θ in [0,2π]that satisfy the equation sin2θ−cosθ14

$The number of values of θ i n [0, 2 π] t h a t s a t i s f y t h e e q u a t i o n s i n^{2} θ - c o s θ \frac{1}{4}$