Find the value of sin x in the interval 0 - x - satisfying the equation 10 sin 2 x-sin x-2-0-

The correct option is B 1/2 10sin2x sin x 2 = 0 Or, 10sin2x 5 sin x + 4 sin x 2 = 0 Or, 5sinx (2 sin x 1) + 2(2sinx 1) = 0 Or, (5sin x + 2) (2 sin ...

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Standard IX

Mathematics

Introduction to Trigonometry

Find the valu...

Question

Find the value of sin x in the interval 0 ≤ x ≤ π, satisfying the equation 10sin²x - sinx - 2 = 0.

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Solution

The correct option is B
1/2

10sin²x - sin x - 2 = 0

Or, 10sin²x - 5 sin x + 4 sin x - 2 = 0

Or, 5sinx (2 sin x -1) + 2(2sinx - 1) = 0

Or, (5sin x + 2) (2 sin x- 1) = 0

Or, sin x = -2/5, 1/2

Since in interval 0 ≤ x ≤ π , sin x is always positive.

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Find the value of sin x in the interval 0 ≤ x ≤ π, satisfying the equation 10sin2x - sinx - 2 = 0.

The correct option is B 1/2 10sin2x - sin x - 2 = 0 Or, 10sin2x - 5 sin x + 4 sin x - 2 = 0 Or, 5sinx (2 sin x -1) + 2(2sinx - 1) = 0 Or, (5sin x + 2) (2 sin x- 1) = 0 Or, sin x = -2/5, 1/2 Since in interval 0 ≤ x ≤ π , sin x is always positive.

Find the value of sin x in the interval 0 ≤ x ≤ π, satisfying the equation 10sin²x - sinx - 2 = 0.

The correct option is B
1/2

10sin²x - sin x - 2 = 0

Or, 10sin²x - 5 sin x + 4 sin x - 2 = 0

Or, 5sinx (2 sin x -1) + 2(2sinx - 1) = 0

Or, (5sin x + 2) (2 sin x- 1) = 0

Or, sin x = -2/5, 1/2

Since in interval 0 ≤ x ≤ π , sin x is always positive.