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Solve for x

 

3cos (x) +2(sin)^2 (x)=0

 Mar 11, 2020
 #1
avatar+12298 
+1

Solve for x

3cos (x) +2(sin)^2 (x)=0

laugh

 Mar 11, 2020
 #2
avatar+25480 
+2

Solve for x
\(3\cos(x) +2\sin^2 (x)=0\)

 

\(\begin{array}{|rcll|} \hline 3\cos(x) +2\sin^2(x) &=& 0 \\ 3\cos(x) &=& -2\sin^2(x) \quad &| \quad : (-2) \\ -1.5\cos(x) &=& \sin^2(x) \\ \hline \end{array}\)

 

Substitute:

\(\begin{array}{|rclcrcl|} \hline t=\tan\left(\dfrac{x}{2}\right) \\ \hline \cos(x) &=& \dfrac{1-t^2}{1+t^2} && \sin(x) &=& \dfrac{2t}{1+t^2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline -1.5\cos(x) &=& \sin^2(x) \quad | \quad \cos(x) = \dfrac{1-t^2}{1+t^2},\ \sin(x) = \dfrac{2t}{1+t^2} \\ -1.5\dfrac{(1-t^2)}{(1+t^2)} &=& \dfrac{4t^2}{(1+t^2)^2} \\\\ -1.5 (1-t^2) &=& \dfrac{4t^2}{ 1+t^2 } \\\\ -1.5 (1-t^2)(1+t^2) &=& 4t^2 \\ -1.5 (1-t^4) &=& 4t^2 \\ -1.5+1.5t^4 &=& 4t^2 \\ \mathbf{ 1.5t^4 -4t^2-1.5} &=& \mathbf{0} \\\\ t^2 &=& \dfrac{4\pm \sqrt{16-4*(1.5)*(-1.5)} }{2*1.5} \\ t^2 &=& \dfrac{4\pm \sqrt{16+9} }{3} \\ t^2 &=& \dfrac{4\pm \sqrt{25} }{3} \\ t^2 &=& \dfrac{4\pm 5 }{3} \\\\ t^2_1 &=& \dfrac{4+5}{3} \\ t^2_1 &=& \dfrac{9}{3} \\ \mathbf{ t^2_1 } &=& \mathbf{3} \\\\ t^2_2 &=& \dfrac{4-5}{3} \\ \mathbf{ t^2_2 } &=& \mathbf{ -\dfrac{1}{3}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcl|rcl|} \hline \mathbf{x=\ ?} && &\mathbf{x=\ ?} \\ \hline \tan\left(\dfrac{x}{2}\right) &=& t_1 & \tan\left(\dfrac{x}{2}\right) &=& t_2 \\ \tan\left(\dfrac{x}{2}\right) &=& \sqrt{3} & \tan\left(\dfrac{x}{2}\right) &=& \sqrt{-\dfrac{1}{3}} \qquad (\text{the root is imaginary !}) \\ \hline \end{array} \)

 

\(\begin{array}{|rcl|rcl|} \hline \tan\left(\dfrac{x}{2}\right) = \sqrt{3} && & \tan\left(\dfrac{x}{2}\right) = \sqrt{3} \\ \hline \tan\left(\dfrac{x}{2}\right) &=& +\sqrt{3} & \tan\left(\dfrac{x}{2}\right) &=& -\sqrt{3} \\ \dfrac{x}{2} &=& \arctan(\sqrt{3})+180^\circ k \quad k\in\mathbb{Z} & \dfrac{x}{2} &=& \arctan(-\sqrt{3})+180^\circ k \quad k\in\mathbb{Z} \\ \dfrac{x}{2} &=& 60^\circ+180^\circ k \quad k\in\mathbb{Z} & \dfrac{x}{2} &=& -60^\circ+180^\circ k \quad k\in\mathbb{Z} \\ \mathbf{x} &=& \mathbf{120^\circ+360^\circ k \quad k\in\mathbb{Z}} & x &=& -120^\circ+360^\circ k \quad k\in\mathbb{Z} \\ & & & x &=& 360^\circ-120^\circ+360^\circ k \quad k\in\mathbb{Z} \\ & & & \mathbf{x} &=& \mathbf{240^\circ+360^\circ k \quad k\in\mathbb{Z}} \\ \hline \end{array}\)

 

laugh

 Mar 12, 2020

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