14.How would you solve this?

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Answer 1 · 2017-12-18T13:44:11

$0 . \frac{π}{2}, \frac{π}{4}, \frac{3 π}{4}$ $color(blue)(0 . pi/2 , pi/4 , (3pi)/4)$

Let $u = 2 x$ $u=2x$

$3 sin (3 u) + 3 sin (u)$ $3sin(3u)+3sin(u)$

Using identity:

$sin (3 x) = - {sin}^{3} (x) + 3 {cos}^{2} (x) sin (x)$ $color(red)(sin(3x)=-sin^3(x)+3cos^2(x)sin(x))$

$3 (- {sin}^{3} (u) + 3 {cos}^{2} (u) sin (u)) + 3 sin (u)$ $3(-sin^3(u)+3cos^2(u)sin(u))+3sin(u)$

Factor:

$3 sin (u) (- {sin}^{2} (u) + 3 {cos}^{2} (u) + 1) = 0$ $3sin(u)(-sin^2(u)+3cos^2(u)+1)=0$

$3 sin (u) = 0$ $3sin(u)=0$

$sin (u) = 0 \Rightarrow u = 0, π$ $sin(u)=0=>u=0, pi$

But $u = 2 x 8888$ $u=2xcolor(white)(8888)$ , so $x = \frac{0}{2}, \frac{π}{2}$ $x=color(blue)(0/2, pi/2)$

$- {sin}^{2} (u) + 3 {cos}^{2} (u) + 1 = 0$ $-sin^2(u)+3cos^2(u)+1=0$

Identity:

$1 - {sin}^{2} (u) = {cos}^{2} (u)$ $color(red)(1-sin^2(u)=cos^2(u))$

${cos}^{2} (u) + 3 {cos}^{2} (u) = 0$ $cos^2(u)+3cos^2(u)=0$

$4 {cos}^{2} (u) = 0$ $4cos^2(u)=0$

${cos}^{2} (u) = 0 \Rightarrow u = \frac{π}{2}, \frac{3 π}{2}$ $cos^2(u)=0=>u=pi/2, (3pi)/2$

But $u = 2 x 8888$ $u=2xcolor(white)(8888)$ , so $x = \frac{π}{4}, \frac{3 π}{4}$ $x=color(blue)(pi/4 , (3pi)/4)$