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