The no.of values of x ∈ [ 0 , 4 π ] satisfying ∣ 3 cos x − sin x ∣ ≥ 2 is

Since the maximum value of 3 cos x − sin x is 2 ∣ 3 cos x − sin x ∣ = 2 only cos ( x + 6 π ) = ± 1 x = 6 5 π , 6 11 π , 6 17 π , 6 23 π

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MathematicsTrigonometric EquationTrigonometric InequationsEasy2 minQB

MathematicsEasyinteger

The no.of values of $x \in [0, 4 π]$ satisfying $∣ 3 cos x - sin x ∣ \geq 2$ is

Answer:

Solution:

Since the maximum value of $3 cos x - sin x$ is 2

$∣ 3 cos x - sin x ∣ = 2$ only $cos (x + \frac{π}{6}) = \pm 1$

$x = \frac{5 π}{6}, \frac{11 π}{6}, \frac{17 π}{6}, \frac{23 π}{6}$

Stream:JEESubject:MathematicsTopic:Trigonometric EquationSubtopic:Trigonometric Inequations

⏱ 2mℹ️ Source: QB

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