Respuesta :
Answer with explanation:
The equation using trigonometric function are whose value we have to evaluate in the interval [0, 2π].
1.→
cos²x +2 cos x=3
→cos²x +2 cos x-3=0
→cos²x +3 cos x-cos x-3=0
→cos x ( cos x +3) -1×(cos x +3)=0
→ (cos x -1)(cos x +3)=0
→cos x -1= 0 ∧ →cos x+3=0
→cos x =1 ∧ → cos x = -3
⇒-1 ≤ cos x ≤ 1
→cos x=1
[tex]\cos x=\cos 0^{\circ}\\\\x=2 n\pi \pm 0^{\circ}\\\\x=2 n\pi[/tex]
where, n is any integer.
→Solution in the Interval [0, 2π] is , x=0, 2 π
2.
2 sin² Ф +7 sin Ф =4
→2 sin² Ф +7 sin Ф -4=0
→2 sin² Ф +8 sin Ф- sin Ф -4=0
→2 sin Ф × (sin Ф +4) -1 × (sin Ф +4)=0
→ (2 sin Ф -1)(sin Ф +4)=0
→ 2 sin Ф -1=0 ∧ →sin Ф +4=0
→sin Ф ≠ -4≡→ ∵ -1 ≤ sin Ф ≤ 1
[tex]\rightarrow \sin \Phi =\frac{1}{2}\\\\\rightarrow\sin \Phi =\sin (\frac{\pi}{6})\\\\ \rightarrow\Phi=n\pi +(-1)^n\frac{\pi}{6}\\\\\rightarrow \Phi=\frac{\pi}{6},\frac{5\pi}{6}[/tex]
Where , n is any Integer.
3.
→tan² x+4=2 sec² x+tan x
→tan² x+4=2(1+tan²x)+tan x
→2 tan²x -tan²x +tan x +2-4=0
→tan²x +tan x -2=0
→tan²x +2 tan x -tan x -2=0
→tan x (tan x +2) -1×(tan x +2)=0
→(tan x-1)(tan x +2)=0
→tan x-1 =0 ∧ tan x +2=0
→tan x=1 ∧ tan x = -2
→tan x=1
[tex]\rightarrow \tan x=\tan (\frac{\pi}{4})\\\\x=n\pi+ \frac{\pi}{4}\\\\x=\frac{\pi}{4},\frac{5\pi}{4}[/tex]
→tan x= -2
[tex]\rightarrow \tan x=2\\\\\rightarrow x=k\pi+ tan^{-1}2\\\\\rightarrow x=tan^{-1}2, \pi +tan^{-1}2[/tex]
where, k and n are integer.
