We have f(x) = x sin x
⇒ f'(x) = x cos x + sin x
⇒ f′′(x) = x (− sin x ) + cos x + cos x = (−x sin x ) + 2 cos x
Now, it is given that f(x) = x sin x satisfies the equation f′′(x) + f(x) + t cos x = 0
⇒ (−x sin x ) + 2 cos x + x sin x + t cos x = 0
⇒ 2 cos x + t cos x = 0
⇒ cos x ( t + 2 ) = 0
⇒ t + 2 = 0
⇒ t = −2