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Solve the following equations for which solution lies in the interval 0° ≤ θ < 360° cos 2x = 1 − 3 sin x

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प्रश्न

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

cos 2x = 1 − 3 sin x

योग
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उत्तर

1 – 2 sin2x = 1 – 3 sinx

2 sin2 x – 3 sin x = 0

sin x(2 sin x – 3) = 0  

= sin x = 0 or 2 sin x – 3 = 0

sin x = 0 or sin x = `3/2`

sin x = `3/2` is not possible since sin x ≤ 1

∴ sin x = 0 = sin 0

The general solution is x = nit,

When n = 0, x = 0 ∉ (0°, 360°)

When n = 1, x = π ∈ (0°, 360°)

When n = 2, x = 2π ∉ (0°, 360°)

∴ The required solutions is x = π

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

APPEARS IN

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

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