Pertanyaan

Himpunan penyelesaian dari tan ( 5 x − 2 5 ∘ ) = − 3 1 3 untuk 0 ∘ ≤ x ≤ 36 0 ∘ adalah ....

Himpunan penyelesaian dari $tan (5 x - 2 5^{\circ}) = - \frac{1}{3} 3$ untuk $0^{\circ} \leq x \leq 36 0^{\circ}$ adalah ....

${3 5^{\circ}, 7 1^{\circ}, 10 7^{\circ}, 14 3^{\circ}, 17 9^{\circ}, 21 5^{\circ}, 25 1^{\circ}, 28 7^{\circ}, 32 3^{\circ}, 35 9^{\circ}}$
${3 5^{\circ}, 6 1^{\circ}, 10 7^{\circ}, 14 5^{\circ}, 17 9^{\circ}, 22 5^{\circ}, 25 1^{\circ}, 28 7^{\circ}, 32 3^{\circ}, 35 9^{\circ}}$
${3 5^{\circ}, 5 1^{\circ}, 10 7^{\circ}, 14 5^{\circ}, 17 9^{\circ}, 22 5^{\circ}, 25 1^{\circ}, 29 7^{\circ}, 32 3^{\circ}, 34 9^{\circ}}$
${4 5^{\circ}, 7 1^{\circ}, 11 7^{\circ}, 14 3^{\circ}, 18 9^{\circ}, 21 5^{\circ}, 26 1^{\circ}, 28 7^{\circ}, 33 4^{\circ}, 35 9^{\circ}}$
${4 5^{\circ}, 6 1^{\circ}, 10 7^{\circ}, 14 3^{\circ}, 17 9^{\circ}, 21 5^{\circ}, 25 1^{\circ}, 29 7^{\circ}, 32 3^{\circ}, 35 9^{\circ}}$

A. Acfreelance

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah A. Ingat tan x = a diubah dahulu menjadi tan x = tan α . Jika tan x = tan α , maka x = α + k ⋅ 18 0 ∘ . Diketahui tan ( 5 x − 5 ∘ ) = − 3 1 3 untuk 0 ∘ ≤ x ≤ 36 0 ∘ , maka tan ( 5 x − 2 5 ∘ ) tan ( 5 x − 2 5 ∘ ) = = = − 3 1 3 ( K . II / IV ) tan ( 18 0 ∘ − 3 0 ∘ ) tan 15 0 ∘ Jadi, himpunan penyelesaiannya adalah { 3 5 ∘ , 7 1 ∘ , 10 7 ∘ , 14 3 ∘ , 17 9 ∘ , 21 5 ∘ , 25 1 ∘ , 28 7 ∘ , 32 3 ∘ , 35 9 ∘ } . Oleh karena itu, jawaban yang benar adalah A.

Jawaban yang benar untuk pertanyaan tersebut adalah A.

Ingat $tan x = a$ diubah dahulu menjadi $tan x = tan α$ .

Jika $tan x = tan α$ , maka $x = α + k \cdot 18 0^{\circ}$ .

Diketahui $tan (5 x - 5^{\circ}) = - \frac{1}{3} 3$ untuk $0^{\circ} \leq x \leq 36 0^{\circ}$ , maka

$tan (5 x - 2 5^{\circ}) tan (5 x - 2 5^{\circ}) = = = - \frac{1}{3} 3 (K . II / IV) tan (18 0^{\circ} - 3 0^{\circ}) tan 15 0^{\circ}$

$\begin{array}{rcl}\mathbf5\mathbf x\boldsymbol-\mathbf{25}\boldsymbol\textdegree&\boldsymbol=&\mathbf{150}\boldsymbol\textdegree\boldsymbol+\boldsymbol k\boldsymbol\cdot\mathbf{180}\boldsymbol\textdegree\\5x&=&175^\circ+k\cdot180^\circ\\x&=&35^\circ+k\cdot36^\circ\end{array}$

$k=0,\;\begin{array}{rcl}&&x\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}&&35\end{array}\begin{array}{rcl}&&\textdegree\end{array}\begin{array}{rcl}&&+\end{array}\begin{array}{rcl}&&0\end{array}\begin{array}{rcl}&&\cdot\end{array}\begin{array}{rcl}&&36\end{array}\begin{array}{rcl}&&\textdegree\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}&&35\end{array}\begin{array}{rcl}&&\textdegree\end{array}\begin{array}{rcl}&&\;\end{array}\begin{array}{rcl}&&(\end{array}\begin{array}{rcl}&&\mathrm{memenuhi}\end{array}\begin{array}{rcl}&&)\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}1,\;x&=&{35^\circ+1\cdot36^\circ=71^\circ\;(\mathrm{memenuhi})}\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}2,\;x&=&{35^\circ+2\cdot36^\circ=107^\circ\;(\mathrm{memenuhi})}\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}3,\;x&=&{35^\circ+3\cdot36^\circ=143^\circ\;(\mathrm{memenuhi})}\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}4,\;x&=&{35^\circ+4\cdot36^\circ=179^\circ\;(\mathrm{memenuhi})}\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}5,\;x&=&{35^\circ+5\cdot36^\circ=215^\circ\;(\mathrm{memenuhi})}\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}6,\;x&=&{35^\circ+6\cdot36^\circ=251^\circ\;(\mathrm{memenuhi})}\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}7,\;x&=&{35^\circ+7\cdot36^\circ=287^\circ\;(\mathrm{memenuhi})}\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}8,\;x&=&{35^\circ+8\cdot36^\circ=323^\circ\;(\mathrm{memenuhi})}\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}9,\;x&=&{35^\circ+9\cdot36^\circ=359^\circ\;(\mathrm{memenuhi})}\end{array}\\\begin{array}{rcl}&&k\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}&&10\end{array}\begin{array}{rcl}&&,\end{array}\begin{array}{rcl}&&\;\end{array}\begin{array}{rcl}&&x\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}&&35\end{array}\begin{array}{rcl}&&\textdegree\end{array}\begin{array}{rcl}&&+\end{array}\begin{array}{rcl}&&10\end{array}\begin{array}{rcl}&&\cdot\end{array}\begin{array}{rcl}&&36\end{array}\begin{array}{rcl}&&\textdegree\end{array}\begin{array}{rcl}&=&\end{array}\begin{array}{rcl}&&395\end{array}\begin{array}{rcl}&&\textdegree\end{array}\begin{array}{rcl}&&\;\end{array}\begin{array}{rcl}&&(\mathrm{tidak}\;\end{array}\begin{array}{rcl}&&\mathrm{memenuhi}\end{array}\begin{array}{rcl}&&)\end{array}$

Jadi, himpunan penyelesaiannya adalah ${3 5^{\circ}, 7 1^{\circ}, 10 7^{\circ}, 14 3^{\circ}, 17 9^{\circ}, 21 5^{\circ}, 25 1^{\circ}, 28 7^{\circ}, 32 3^{\circ}, 35 9^{\circ}}$ .