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Solve the following equations for which solution lies in the interval 0° ≤ θ < 360° sin4x = sin2x

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प्रश्न

Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

sin4x = sin2x

योग
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उत्तर

sin4x – sin2x = 0

sin2 x (sin2 x – 1) = 0

sin2 x [– (1 – sin2 x)] = 0

sin2x × – cos2x = 0

– sin2x cos2x = 0

(sin x cos x)2 = 0

`(1/2 xx 2 sin cos x)^2` = 0

`1/4 sin 2x` = 0

sin 2x = 0

The general solution is

2x = nπ, n ∈ Z

x = `("n"pi)/2`, n ∈ Z

When n = 0, x = `(0 xx pi)/2` = 0 ∉ (0°, 360°)

When n = 1, x = `pi/2` = ∈ (0°, 360°)

When n = 2, x = `(2pi)/2` = π ∈ (0°, 360°)

When n = 3, x = `(3pi)/2` = ∈ (0°, 360°)

When n = 4, x = `(4pi)/2` = 2π ∉ (0°, 360°)

∴ The values of x are `pi/2`, π, `(3pi)/2`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.8 [पृष्ठ १३३]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.8 | Q 2. (i) | पृष्ठ १३३

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