Question

Ifx² - 2x + sin²= 0, then x belongs to​

Answer 1

Answer:

Any digit raised to power to zero equal to 1.

a/b=0

a=0

Answer 2

Question:

If $x^2 - 2x + sin^2 \theta = 0$ , then x belongs to.

Solution:

$\tt x^2 - 2x + sin^2 \theta = 0 \\\\\tt x^2-2x= -sin^2 \theta \\\\ \tt Equate~the~range~of~x^2-2x~and~-sin^2\theta \\\\ \\\\ \tt We~know~that~sin\theta ~lies~in~range~[-1,1 ] ~i.e ~sin \theta \epsilon [-1,1] \\\\ \tt sin^2 \theta \epsilon [0,1] \\\\ \tt- sin^2 \theta \epsilon [-1,0] \\\\ \tt Now, ~the ~range ~of ~x^2-2x~should ~also~ come~ under~ [-1,0] \\\\ \tt - 1<x^2-2x<0 \\\\ \tt \: First, let's~find~x~for~-1<x^2-2x\\\\ x^2-2x>-1~\\\\ \tt \: x^2-2x>-1 \\\\ \tt x^2-2x+1>0 \\\\ \tt x^2-x-x+1>0$$\\\\ \tt (x-1)^2>0 \\\\ \tt x \epsilon R. \\\\ \tt Now, ~find~x~for~x^2-2x<0 \\\\ \tt x^2-2x<0 \\\\ \tt x(x-2)<0 \\\\ \tt Use~wavy~curve~to~determine~the~value~of~x, here ~the~critical~points~are~0~and~2 \\\\ \tt Hence, the~value~of~x~for~ \boxed{x<0}~lies~b/w~0~and~2 \\\\ \tt x \epsilon [0,2] \\\\ \tt x \epsilon \: R\cap[0,2] \\\\ \huge \tt x \epsilon[0,2]$