# Solving trigonometric equations

• Apr 21st 2009, 04:06 PM
etheen
Solving trigonometric equations
Solve over Real numbers:
a)5sin^2t+cos^2t-2=0 b)sec^2t+tantt=7 c)root of sint=2sint-1 d)tant-sintcost=0

Solve over [0, 2pi[ :

e)sin^2x-cos^2x=0 f)2tan^2x+3tanx-5=0 g)2sin^2x-sinx-1=0 h)6cos^2x+5cosx=0

solve:
2(cos(3x-pi)+2=1 "x" element of reals

if "x" element [0 degrees,180degrees] solve 4sin2(x-5)=1

Find the exact value:
cosx=-1/3 pi/2<0<pi

evaluate
sin2x tan2x
• Apr 21st 2009, 08:15 PM
Soroban
Hello, etheen!

I'll do a few of them . . .

Quote:

Solve over Real numbers:

$\displaystyle a)\;5\sin^2\!t+\cos^2\!t-2\:=\:0$

Replace $\displaystyle \cos^2\!t$ with $\displaystyle 1 - \sin^2\!t$

$\displaystyle 5\sin^2\!t + (1 - \sin^2\!t) - 2 \:=\:0 \quad\Rightarrow\quad 4\sin^2\!t - 1 \:=\:0 \quad\Rightarrow\quad 4\sin^2\!t \:=\:1$

. . $\displaystyle \sin^2\!t \:=\:\tfrac{1}{4} \quad\Rightarrow\quad \sin t \:=\:\pm\tfrac{1}{2} \quad\Rightarrow\quad t \;=\;\begin{Bmatrix}\frac{\pi}{6} + 2\pi n \\ \\[-3mm] \frac{5\pi}{6} + 2\pi n \end{Bmatrix}$

Quote:

$\displaystyle b)\;\sec^2\!t+\tan t\:=\:7$
Replace $\displaystyle \sec^2\!t$ with $\displaystyle \tan^2\!t + 1$

$\displaystyle (\tan^2\!t + 1) + \tan t \:=\:7 \quad\Rightarrow\quad \tan^2\!t + \tan t - 6 \:=\:0 \quad\Rightarrow\quad (\tan t - 2)(\tan t + 3) \:=\:0$

We have: .$\displaystyle \begin{array}{cccccccccc} \tan t - 2 \:=\:0 & \Rightarrow & \tan t \:=\:2 & \Rightarrow & t \:=\:\arctan(2) + \pi n \\ \tan t + 3 \:=\:0 & \Rightarrow & \tan t \:=\:\text{-}3 & \Rightarrow & t \:=\:\arctan(\text{-}3) + \pi n \end{array}$

Quote:

$\displaystyle c)\;\sqrt{\sin t}\:=\:2\sin t-1$

Square both sides: .$\displaystyle \left(\sqrt{\sin t}\right)^2 \:=\:(2\sin t - 1)^2 \quad\Rightarrow\quad \sin t \:=\:4\sin^2\!t - 4\sin t + 1$

. . $\displaystyle 4\sin^2\!t - 5\sin t + 1 \:=\:0 \quad\Rightarrow\quad (\sin t - 1)(4\sin t - 1) \:=\:0$

We have: .$\displaystyle \begin{array}{ccccccc} \sin t - 1 \:=\:0 & \Rightarrow & \sin t \:=\:1 & \Rightarrow & t \:=\:\frac{\pi}{2} + 2\pi n \\ \\[-3mm] 4\sin t - 1 \:=\:0 & \Rightarrow & \sin t \:=\:\frac{1}{4} & \Rightarrow& t \:=\:\arcsin\left(\frac{1}{4}\right) + 2\pi n \end{array}$

Quote:

$\displaystyle d)\;\tan t-\sin t\cos t\:=\:0$

We have: .$\displaystyle \frac{\sin t}{\cos t} - \frac{\sin t \cos^2\!t}{\cos t} \:=\:0 \quad\Rightarrow\quad \frac{\sin t - \sin t\cos t}{\cos t} \:=\:0$

. . $\displaystyle \frac{\sin t(1 - \cos^2\!t)}{\cos t} \:=\:0 \quad\Rightarrow\quad \frac{\sin t\cdot\sin^2\!t}{\cos t} \:=\:0 \quad\Rightarrow\quad \frac{\sin^3\!t}{\cos t} \:=\:0$

. . $\displaystyle \sin^3\!t \:=\:0 \quad\Rightarrow\quad \sin t \:=\:0 \quad\Rightarrow\quad t \:=\:\pi n$

• Apr 22nd 2009, 12:52 PM
etheen
hm..i see, I see. These are the harder ones. I'm going to do a few more and ask if I am stuck...Thank you so much :}