If lo g 0.5 sin x = 1 − lo g 0.5 cos x then number of values of x ∈ [ − 2 π , 2 π ] is

lo g 0.5 sin x = 1 − lo g 0.5 cos x x ∈ [ − 2 π , 2 π ] sin x > 0 an d cos x > 0 sin x cos x = 2 1 sin 2 x = 1 2 x ∈ [ − 4 π , 4 π ] ⇒ 4 solutions

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MathematicsTrigonometric EquationTrigonometric EquationMedium2 minQB

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If $lo g_{0.5} sin x = 1 - lo g_{0.5} cos x$ then number of values of $x \in [- 2 π, 2 π]$ is

Answer:

Solution:

$lo g_{0.5} sin x = 1 - lo g_{0.5} cos x$

$x \in [- 2 π, 2 π] sin x > 0 an d cos x > 0 sin x cos x = \frac{1}{2} sin 2 x = 12 x \in [- 4 π, 4 π] \Rightarrow$

4 solutions

Stream:JEESubject:MathematicsTopic:Trigonometric EquationSubtopic:Trigonometric Equation

⏱ 2mℹ️ Source: QB

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