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The number of values of xx in the interval [0,5π][0,5π] satisfying the equation 3sin2x−7sinx+2=03sin2⁡x-7sin⁡x+2=0 is/are

Respuesta :

Answer:

6

Step-by-step explanation:

Given, 3sin2x−7sinx+2=03sin2⁡x-7sin⁡x+2=0

⇒(3sinx−1)(sinx−2)=0⇒3sin⁡x-1sin⁡x-2=0

⇒sinx=13 or 2⇒sin⁡x=13 or 2

⇒sinx=13    [∵sinx≠2]⇒sin⁡x=13    [∵sin⁡x≠2]

Let  sinα=13,0<α<π2,sinα=13,0<α<π2, then sinx=sinαsinx=sinα

now x=nπ+(−1)nα(n∈I)x=nπ+(−1)nα(n∈I)

⇒x=α,π−α,2π+α,3π−α,4π+α,5π−α⇒x=α,π−α,2π+α,3π−α,4π+α,5π−α Are the solution in [0,5π][0,5π]

Hence, required number of solutions are 6

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