Could someone please demonstrate to me how to solve these equations:
a) arccos(x) + arccos(x*sqrt(3)) = pi/2
b) 2arcsin(x*sqrt(6)) + arcsin(4x) = pi/2</SPAN>
see this thread ... solution process is the same.
http://www.mathhelpforum.com/math-he...ease-help.html
Hello, Sunyata!
Let: .$\displaystyle \begin{array}{ccccccccccc}\alpha \;=\; \arccos(x) &\Longrightarrow & \cos\alpha \;=\; x & \Longrightarrow & \sin\alpha \;=\; \sqrt{1-x^2} \\$\displaystyle (a)\;\;\arccos(x) + \arccos(x\sqrt{3}) \:=\:\frac{\pi}{2}$
\beta \;=\; \arccos(x\sqrt{3}) & \Longrightarrow & \cos\beta \;=\; x\sqrt{3} & \Longrightarrow & \sin\beta \;=\; \sqrt{1-3x^2}
\end{array}$ .[1]
The equation becomes: .$\displaystyle \alpha + \beta \;=\;\frac{\pi}{2}$
Take the cosine: .$\displaystyle \cos(\alpha + \beta) \;=\;\cos\left(\tfrac{\pi}{2}\right) $
. . which becomes: .$\displaystyle \cos\alpha\cos\beta - \sin\alpha\sin\beta \;=\;0$
Substitute [1]: .$\displaystyle (x)(x\sqrt{3}) - (\sqrt{1-x^2})(\sqrt{1-3x^2}) \:=\:0 $
. . . . . . $\displaystyle x^2\sqrt{3} - \sqrt{1-4x^2 + 3x^4} \:=\:0 \quad\Rightarrow\quad x^2\sqrt{3} \:=\:\sqrt{1 - 4x^2 + 3x^4} $
Square both sides: .$\displaystyle 3x^4 \:=\:1 - 4x^2 + 3x^4 \quad\Rightarrow\quad 4x^2 \:=\:1 \quad\Rightarrow\quad x^2 \:=\:\frac{1}{4}$
. . Therefore: .$\displaystyle x \;=\;\pm\frac{1}{2}$