Solve the Trig equation
- Thread starter flakine
- Start date
I think I got it:
(sin x+1)/cos x=sqrt3
square both sides:
(sin X+1)^2/Cos^2 x=(sqrt3)^2
(sin^2 x + 2sin x + 1)/(1-sin^2 x) =3
(sin^2 x + 2sin x + 1)=3(1-sin^2 x)
(sin^2 x + 2sin x + 1)= 3-3sin^2 x
4sin^2 x + 2sin x -2=0
(2sin x-1)(sin x +1)=0
sin x=1/2 & sin x=-1
x=pie/6 & x=3pie/2
Is this correct? Someone help please!!
(sin x+1)/cos x=sqrt3
square both sides:
(sin X+1)^2/Cos^2 x=(sqrt3)^2
(sin^2 x + 2sin x + 1)/(1-sin^2 x) =3
(sin^2 x + 2sin x + 1)=3(1-sin^2 x)
(sin^2 x + 2sin x + 1)= 3-3sin^2 x
4sin^2 x + 2sin x -2=0
(2sin x-1)(sin x +1)=0
sin x=1/2 & sin x=-1
x=pie/6 & x=3pie/2
Is this correct? Someone help please!!
Hello, flakine!
Use the same technique for the first one . . .
We have: .\(\displaystyle \tan x\,+\,\sec x\:=\:\sqrt{3}\)
Then: .\(\displaystyle \sec x\:=\:\sqrt{3}\,-\,\tan x\)
Square: .\(\displaystyle \sec^2x\:=\:3\,-\,2\sqrt{3}\cdot\tan x\,+\,\tan^2x\)
. . \(\displaystyle \tan^2x\,+\,1\:=\:3\,-\,2\sqrt{3}\cdot\tan x\,+\,\tan^2x\)
which simplifies to: .\(\displaystyle 2\sqrt{3}\cdot\tan x\:=\:2\;\;\Rightarrow\;\;\tan x\,=\,\frac{1}{\sqrt{3}}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{6},\,\frac{7\pi}{6}\)
And, as tkhunny cautioned, check these answers.
Use the same technique for the first one . . .
We have: .\(\displaystyle \tan x\,+\,\sec x\:=\:\sqrt{3}\)
Then: .\(\displaystyle \sec x\:=\:\sqrt{3}\,-\,\tan x\)
Square: .\(\displaystyle \sec^2x\:=\:3\,-\,2\sqrt{3}\cdot\tan x\,+\,\tan^2x\)
. . \(\displaystyle \tan^2x\,+\,1\:=\:3\,-\,2\sqrt{3}\cdot\tan x\,+\,\tan^2x\)
which simplifies to: .\(\displaystyle 2\sqrt{3}\cdot\tan x\:=\:2\;\;\Rightarrow\;\;\tan x\,=\,\frac{1}{\sqrt{3}}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{6},\,\frac{7\pi}{6}\)
And, as tkhunny cautioned, check these answers.