Example 1:
There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.
We know that the
Example 2:
![$2\sin\left( 3x\right) -1=0$](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u-j-aYlFA3SF3iTKgn3xcA18fmqsjJ_KS28SorjYpL0onry9XPQU6riuMDspb0MsCFbEr-k79R22qtaOLV9qm_OeKZat8kfj5C54MqHQmGJ8-FDfXl9p_FzQDR=s0-d)
There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.
![\begin{eqnarray*}&& \\
2\sin \left( 3x\right) -1 &=&0 \\
&& \\
\sin \left( 3x\right) &=&\displaystyle \frac{1}{2} \\
&& \\
&&
\end{eqnarray*}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vx52YLhzLXlh6p3zG8Z-8nzPB6pYMivJTJ1c0fpew25vFiOrEdBA6zfe0WHxWF2_0hVi4bYtqI3zpXuNsG5JuOjK2Qc-NKRfIBHBYjTcpOERUH1ReJ2H90tyY=s0-d)
If we restriction the domain of the sine function to
, we can use the inverse sine function to solve for reference angle 3x and then x.
![\begin{eqnarray*}&& \\
\sin \left( 3x\right) &=&\displaystyle \frac{1}{2} \\
&...
...&=&\sin ^{-1}\left( \displaystyle \frac{1}{2
}\right) \\
&& \\
\end{eqnarray*}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vQmCstzzyjAw0OMZtyIvbbn5pzd7YeXHt1FZ8VshXzHiQFCuRiyWGo8J_Oytl5nPlkRTsxYwaXJ7guKESM4euKUicSZGya7lHjqrrq_mTrP-sMe6693Xes5pKT=s0-d)
![\begin{eqnarray*}3x &=&\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \\
&& ...
...\
x &\approx &0.174532925\ \mbox{ radians }\\
&& \\
&& \\
&&
\end{eqnarray*}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vwfwofWdRy5W9suw9zWRXtLfntRdYMmUtjlM6BAN_Dy20IhqEaRh4do0CBFDol9MtmitA1q0wEepHZBX9CUNmU2P3xC4-KR3kDVBtkKoJ66t622i2Ub5rt_3a8=s0-d)
We know that the
e function is positive in the first and the second quadrant. Therefore two of the solutions are the angle 3x that terminates in the first quadrant and the angle
that terminates in the second quadrant. We have already solved for 3x.
![\begin{eqnarray*}&& \\
\sin \left( \pi -3x\right) &=&\displaystyle \frac{1}{2} ...
...-3x &=&\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \\
&&
\end{eqnarray*}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vhU32fHpyw6hxIDNv7-npVfAG5P0qbiNrsX32YAh-cpSQ405LOZug4bs1LVEXXFjCXrNq_6DpCyv-Vlzzg8_IiP0gSE0Uvm6CbcZW5waNYQ7TiguWcYUssw6Hu=s0-d)
The solutions are
and ![$
x=\displaystyle \frac{\pi }{3}-\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) .\bigskip
\bigskip\bigskip $](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s949rjPPbHAmNFgySaUjW3pW_YMTQ-JQDoilxwBS-JvkUKTQqNclfM5wgBxR1RBC-2cwxrJkEM4Onr1ZmPMF-5vuKiVRjrF8xg8iLyRCzT58JYawRc3z9OG5uf9g=s0-d)
Example 3.
Solve the equation
for 0 ≤ θ < 2π.
Example 4.
Solve the equation 6 sin2θ − sin θ − 1 = 0 for 0 ≤ θ < 2π.
Solve the equation tan 2θ − cot 2θ = 0 for 0 ≤ θ < 2π.
![image](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tdIsQEguDs2kA7hpfO5vwNBoeitqUefcVr2Ux1-rp8lPPwIGtgCQCix2CNnPcszDbbtPdVuPsqF1KCLq0Vh-_6z0Y_rEMXcUGWOOctD-IDr6Ys6JOfzTJiCQ=s0-d)
tan2 2θ = 1
Since 0 ≤ θ < 2π , we need to consider values of 2θ such that 0 ≤ 2θ < 4π. Hence, solving the above equation, we have:
There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.
If we restriction the domain of the sine function to
We know that the
Example 3.
Solve the equation
Solving for cos θ gives us:
If
, then the reference angle is α = 1.3181.
So for
, we have θ in the first and 4th quadrants. So
θ = 1.3181 or 4.9651
For
, we have θ in the 2nd and 3rd quadrants. So
θ = 1.8235 or 4.4597
So θ= 1.3181, 1.8235, 4.4597 or 4.9651 radians.
Solve the equation 6 sin2θ − sin θ − 1 = 0 for 0 ≤ θ < 2π.
Factoring the LHS:
6 sin2 θ − sin θ − 1 = 0
(2 sin θ − 1)(3 sin θ + 1) = 0
So either
2 sin θ − 1 = 0Example 5.
sin θ = 1/2
θ will be in 1st and 2nd quadrants.
θ = 0.52360, 2.6180 (ie π/6, 5π/6)
Solve the equation tan 2θ − cot 2θ = 0 for 0 ≤ θ < 2π.
tan2 2θ = 1
tan 2θ = ± 1
Since 0 ≤ θ < 2π , we need to consider values of 2θ such that 0 ≤ 2θ < 4π. Hence, solving the above equation, we have:
So
* For More : http://tutorial.math.lamar.edu/Classes/CalcI/TrigEquations.aspx
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