Đáp án:
Giải thích các bước giải:
c) `tan\ (2x+\frac{\pi}{4})=-\frac{\sqrt{3}}{3}`
`⇔ tan\ (2x+\frac{\pi}{4})=-\frac{\sqrt{3}}{3}`
`⇔ tan\ (2x+\frac{\pi}{4})=tan\ (-\frac{\pi}{6})`
`⇔ 2x+\frac{\pi}{4}=-\frac{\pi}{6}+k\pi\ (k \in \mathbb{Z})`
`⇔ 2x=-\frac{5\pi}{12}+k\pi\ (k \in \mathbb{Z})`
`⇔ x=-\frac{5\pi}{24}+k\frac{\pi}{2}\ (k \in \mathbb{Z})`
`0 < x < \pi`
`⇔ 0 < -\frac{5\pi}{24}+k\frac{\pi}{2} < \pi`
`⇔ 0 < -\frac{5}{24}+k\frac{1}{2} < 1`
`⇔ 5/24 < 1/2 k < 29/24`
`⇔ 5/12 < k < 29/12`
Mà `k \in \mathbb{Z} ⇒ k = {1;2}`
`⇒` Có 2 nghiệm
d) `sin\ (2x-15^{0})=\frac{\sqrt{2}}{2}`
`⇔ sin\ (2x-15^{0})=sin\ 45^{0}`
`⇔` \(\left[ \begin{array}{l}2x-15^{0}=45^{0}+k360^{0}\ (k \in \mathbb{Z})\\2x-15^{0}=180^{0}-45^{0}+k360^{0}\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x-15^{0}=45^{0}+k360^{0}\ (k \in \mathbb{Z})\\2x-15^{0}=135^{0}+k360^{0}\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=30^{0}+k180^{0}\ (k \in \mathbb{Z})\\x=75^{0}+k180^{0}\ (k \in \mathbb{Z})\end{array} \right.\)
`-120^{0} < x < 90^{0}`
`⇔ -120^{0} < 30^{0}+k180^{0} < 90^{0}`
`⇔ -2/3 < 1/6+k<1/2`
`⇔ -5/6 < k < 1/3`
Mà `k \in \mathbb{Z} ⇒ k=0`
`⇒` Có 1 nghiệm
`-120^{0} < x < 90^{0}`
`⇔ -120^{0} < 75^{0}+k180^{0}< 90^{0}`
`⇔ -2/3 < 3/4+k<1/2`
`⇔ -17/12 < k < -1/4`
Mà `k \in \mathbb{Z} ⇒ k = -1`
`⇒` Có 1 nghiệm
