Question

For m, $n\in N$,
Let A(n) = $16\left({\cos }^n\theta -{\sin }^n\theta \right)$ and B(m) = $16\left({\cos }^m\theta +{\sin }^m\theta \right)$ and ${\log }_{10}2=0.3010$
Number of solution for $\theta \in \left[0,2\pi \right]$ such that $A\left(4\right)=B\left(3\right)$ is equal to

• A. $2$
• B. $6$
• C. $4$
• D. $3$

Answer 1

The correct option is A $2$

$A\left(n\right)=16(\cos {}^n\theta -\sin {}^n\theta )B\left(m\right)=16(\cos {}^m\theta +\sin {}^m\theta )A\left(4\right)=B\left(3\right)\Rightarrow 16(\cos {}^4\theta -\sin {}^4\theta )=6(\cos {}^3\theta +\sin {}^3\theta )\Rightarrow 16(\cos {}^2\theta +\sin {}^2\theta )(\cos {}^2\theta -\sin {}^2\theta )=16(\cos {}^3\theta +\sin {}^3\theta )\Rightarrow (\cos \theta +\sin \theta )(\cos \theta -\sin \theta )=(\cos \theta +\sin \theta )(\cos {}^2\theta +\sin {}^2\theta -\sin \theta \cos \theta )\Rightarrow \cos \theta -\sin \theta =1-\sin \theta \cos \theta \Rightarrow \cos \theta -\sin \theta -1+\sin \theta \cos \theta =0\Rightarrow \cos \theta (1+\sin \theta )-1(1+\sin \theta )=0\therefore 1+\sin \theta =0,\cos \theta -1=01+\sin \theta =0\Rightarrow \sin \theta =-1,\theta =\frac{3\pi }{2}\cos \theta =1\Rightarrow \theta =2\pi$

$\therefore 2$ solutions are possible.