Question

If ${\sin }^2\left(\frac{x+3}{2}\right)+{\cos }^2\left(\frac{2x}{3}\right)={\tan }^{2}45$, then $x=...................$

• A. $9$
• B. $-15$
• C. $\frac{15}{7}$
• D. $15$

Answer 1

The correct option is A $9$

Given

${\sin }^2\left(\frac{x+3}{2}\right)+{\cos }^2\left(\frac{2x}{3}\right)={\tan }^{2}45$

$⟹{\sin }^2\left(\frac{x+3}{2}\right)+{\cos }^2\left(\frac{2x}{3}\right)=1$

$⟹{\sin }^2\left(\frac{x+3}{2}\right)=1-{\cos }^2\left(\frac{2x}{3}\right)$

$⟹{\sin }^2\left(\frac{x+3}{2}\right)={\sin }^2\left(\frac{2x}{3}\right)$

$({\sin }^{2}x+{\cos }^{2}x=1)$

$⟹\left(\frac{x+3}{2}\right)=\left(\frac{2x}{3}\right)$

$⟹3x+9=4x$

$⟹x=9$