Trigonometrical Equations 3 Question 4

4. The number of values of x in the interval [0,5π] satisfying the equation 3sin2⁡x−7sin⁡x+2=0 is

(1998, 2M)

(a) 0

(b) 5

(c) 6

(d) 10

Show Answer

Answer:

Correct Answer: 4. (c)

Solution:

  1. Given, 3sin2⁡x−7sin⁡x+2=0

⇒3sin2⁡x−6sin⁡x−sin⁡x+2=0

⇒3sin⁡x(sin⁡x−2)−1(sin⁡x−2)=0

⇒(3sin⁡x−1)(sin⁡x−2)=0

⇒sin⁡x=13[∵sin⁡x=2 is rejected ]

⇒x=nπ+(−1)nsin−1⁡13,n∈I

For 0≤n≤5,x∈[0,5π]

There are six values of x∈[0,5π] which satisfy the equation 3sin2⁡x−7sin⁡x+2=0.

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