Total number of solutions for the equation ${sin}^{4} x + {cos}^{4} x = sin x cos x, x \in [0, 2 π]$ is

2

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4

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6

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3

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Solution

The correct option is A $2$
${sin}^{4} x + {cos}^{4} x = sin x cos x \Rightarrow ({sin}^{2} x + {cos}^{2} x)^{2} - 2 {sin}^{2} x {cos}^{2} x = sin x cos x \Rightarrow 1 - \frac{{sin}^{2} 2 x}{2} = \frac{sin 2 x}{2} \Rightarrow {sin}^{2} 2 x + sin 2 x - 2 = 0 \Rightarrow (sin 2 x + 2) (sin 2 x - 1) = 0 \Rightarrow sin 2 x = 1 (∵ - 1 \leq sin 2 x \leq 1) \Rightarrow 2 x = (4 n + 1) \frac{π}{2}, n \in Z \Rightarrow x = (4 n + 1) \frac{π}{4}, n \in Z$
So the solution's in $[0, 2 π]$ are $x = \frac{π}{4}, \frac{5 π}{4}$

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Similar questions

Q. The total number of solution of the equation

{sin}^{4} x + {cos}^{4} x = sin x cos x

[0, 2 π]

is :

Q. The total number of solution of

{sin}^{4} x + {cos}^{4} x = sin x cos x

[0, 2 π)

is equal to:

Q. The number of solutions to the equation

{cos}^{4} x + \frac{1}{c o s^{2} x} = s i n^{4} x + \frac{1}{s i n^{2} x}

in the interval

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?

Q. Solve the following equation:

{sin}^{4} x + {cos}^{4} x = sin x cos x .

The number of distinct solutions of the equation
$\frac{5}{4} c o s^{2} 2 x + c o s^{4} x + s i n^{4} x + c o s^{6} x + s i n^{6} x = 2$
In the interval $[0, 2 π]$ is ___

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Solving Simultaneous Trigonometric Equations

Standard XII Mathematics

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Total number of solutions for the equation sin4x+cos4x=sinxcosx ,x∈[0,2π] is

The correct option is A 2sin4x+cos4x=sinxcosx⇒(sin2x+cos2x)2−2sin2xcos2x=sinxcosx⇒1−sin22x2=sin2x2⇒sin22x+sin2x−2=0⇒(sin2x+2)(sin2x−1)=0⇒sin2x=1 (∵−1≤sin2x≤1)⇒2x=(4n+1)π2, n∈Z⇒x=(4n+1)π4, n∈Z So the solution's in [0,2π] are x=π4,5π4

Total number of solutions for the equation ${sin}^{4} x + {cos}^{4} x = sin x cos x, x \in [0, 2 π]$ is