Pertanyaan

Untuk 0 < x < π , jika { x ∈ R ∣ a < x < b } adalah himpunan penyelesaian dari 2 cos ( cos x − sin x ) + tan 2 x < sec 2 x maka b − a = ....

Untuk $0 < x < π$ , jika ${x \in R ∣ a < x < b}$ adalah himpunan penyelesaian dari $2 cos (cos x - sin x) + tan^{2} x < sec^{2} x$ maka $b - a = ....$

$\frac{2 π}{8}$
$\frac{3 π}{8}$
$\frac{4 π}{8}$
$\frac{6 π}{8}$
$π$

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S. Nur

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Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Ingat kembali: se c 2 x − tan 2 x sin 2 x cos 2 x sin ( 9 0 ∘ − A ) sin A − sin B sin x = = = = = = 1 2 sin x cos x ⎩ ⎨ ⎧ cos 2 x − sin 2 x 1 − 2 sin 2 x 2 cos 2 x − 1 cos A 2 cos 2 1 ( A + B ) sin 2 1 ( A − B ) sin α { x = α + k ⋅ 36 0 ∘ x = ( 18 0 ∘ − α ) + k ⋅ 36 0 ∘ Maka: 2 cos ( cos x − sin x ) + tan 2 x 2 cos ( cos x − sin x ) 2 cos 2 x − 2 sin x cos x 2 cos 2 x − sin 2 x − 1 2 cos 2 x − 1 − sin 2 x cos 2 x − sin 2 x sin ( 9 0 ∘ − 2 x ) − sin 2 x 2 cos 2 1 ( 9 0 ∘ − 2 x + 2 x ) sin 2 1 ( 9 0 ∘ − 2 x − 2 x ) 2 cos 4 5 ∘ sin ( 4 5 ∘ − 2 x ) 2 ⋅ 2 1 2 ⋅ sin ( 4 5 ∘ − 2 x ) 2 sin ( 4 5 ∘ − 2 x ) sin ( 4 5 ∘ − 2 x ) sin ( 4 5 ∘ − 2 x ) < < < < < < < < < < < < < sec 2 x sec 2 x − tan 2 x 1 0 0 0 0 0 0 0 0 0 sin 0 ∘ Persamaan (1) 4 5 ∘ − 2 x − 2 x x = = = 0 ∘ + k ⋅ 36 0 ∘ − 4 5 ∘ + k ⋅ 36 0 ∘ 22 , 5 ∘ − k ⋅ 9 0 ∘ k = − 1 → x = 22 , 5 ∘ − ( − 1 ) ⋅ 9 0 ∘ = 22 , 5 ∘ + 9 0 ∘ = 112 , 5 ∘ k = 0 → x = 22 , 5 ∘ − 0 ⋅ 9 0 ∘ = 22 , 5 ∘ + 0 = 22 , 5 ∘ Persamaan (2) 4 5 ∘ − 2 x 4 5 ∘ − 2 x − 2 x x = = = = ( 18 0 ∘ − 0 ∘ ) + k ⋅ 36 0 ∘ 18 0 ∘ + k ⋅ 36 0 ∘ 13 5 ∘ + k ⋅ 36 0 ∘ − 67 , 5 ∘ − k ⋅ 9 0 ∘ k = − 1 → x = − 67 , 5 ∘ − ( − 1 ) ⋅ 9 0 ∘ = − 67 , 5 ∘ + 9 0 ∘ = 22 , 5 ∘ k = − 2 → x = − 67 , 5 ∘ − ( − 2 ) ⋅ 9 0 ∘ = − 67 , 5 ∘ + 18 0 ∘ = 112 , 5 ∘ Diperoleh niali a = 22 , 5 ∘ dan b = 112 , 5 ∘ . Dengan demikian: b − a = = = = = 112 , 5 ∘ − 22 , 5 ∘ 9 0 ∘ 9 0 ∘ × 18 0 ∘ π 2 1 π 2 4 π Jadi, jawaban yang benar adalah C.

Ingat kembali:

$se c^{2} x - tan^{2} x sin 2 x cos 2 x sin (9 0^{\circ} - A) sin A - sin B sin x = = = = = = 1 2 sin x cos x ⎩ ⎨ ⎧ cos^{2} x - sin^{2} x 1 - 2 sin^{2} x 2 cos^{2} x - 1 cos A 2 cos \frac{1}{2} (A + B) sin \frac{1}{2} (A - B) sin α {x = α + k \cdot 36 0^{\circ} x = (18 0^{\circ} - α) + k \cdot 36 0^{\circ}$

Maka:

$2 cos (cos x - sin x) + tan^{2} x 2 cos (cos x - sin x) 2 cos^{2} x - 2 sin x cos x 2 cos^{2} x - sin 2 x - 1 2 cos^{2} x - 1 - sin 2 x cos 2 x - sin 2 x sin (9 0^{\circ} - 2 x) - sin 2 x 2 cos \frac{1}{2} (9 0^{\circ} - 2 x + 2 x) sin \frac{1}{2} (9 0^{\circ} - 2 x - 2 x) 2 cos 4 5^{\circ} sin (4 5^{\circ} - 2 x) 2 \cdot \frac{1}{2} 2 \cdot sin (4 5^{\circ} - 2 x) 2 sin (4 5^{\circ} - 2 x) sin (4 5^{\circ} - 2 x) sin (4 5^{\circ} - 2 x) < < < < < < < < < < < < < sec^{2} x sec^{2} x - tan^{2} x 1000000000 sin 0^{\circ}$

Persamaan (1)

$4 5^{\circ} - 2 x - 2 x x = = = 0^{\circ} + k \cdot 36 0^{\circ} - 4 5^{\circ} + k \cdot 36 0^{\circ} 22, 5^{\circ} - k \cdot 9 0^{\circ}$

$k = - 1 \to x = 22, 5^{\circ} - (- 1) \cdot 9 0^{\circ} = 22, 5^{\circ} + 9 0^{\circ} = 112, 5^{\circ} k = 0 \to x = 22, 5^{\circ} - 0 \cdot 9 0^{\circ} = 22, 5^{\circ} + 0 = 22, 5^{\circ}$

Persamaan (2)

$4 5^{\circ} - 2 x 4 5^{\circ} - 2 x - 2 x x = = = = (18 0^{\circ} - 0^{\circ}) + k \cdot 36 0^{\circ} 18 0^{\circ} + k \cdot 36 0^{\circ} 13 5^{\circ} + k \cdot 36 0^{\circ} - 67, 5^{\circ} - k \cdot 9 0^{\circ}$

$k = - 1 \to x = - 67, 5^{\circ} - (- 1) \cdot 9 0^{\circ} = - 67, 5^{\circ} + 9 0^{\circ} = 22, 5^{\circ} k = - 2 \to x = - 67, 5^{\circ} - (- 2) \cdot 9 0^{\circ} = - 67, 5^{\circ} + 18 0^{\circ} = 112, 5^{\circ}$