Q: the equation , sin^( 2) (x)−sin(mx)cos^2 (x)=1 has four distinct roots in ( 0 , 2π ) find the values of , m . (m ∈ N )


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Question Number 201263 by mnjuly1970 last updated on 02/Dec/23

$Q : t h e e q u a t i o n, {s i n}^{2} (x) - s i n (m x) c o s^{2} (x) = 1$ $h a s f o u r d i s t i n c t r o o t s$ $i n (0, 2 π) f i n d t h e v a l u e s$ $o f, m . (m \in N)$
	Answered by MM42 last updated on 03/Dec/23

	$s i n m x \times {c o s}^{2} x + 1 - {s i n}^{2} x = 0$ ${c o s}^{2} x (s i n m x + 1) = 0$ ${c o s}^{2} x = 0 \Rightarrow x = \frac{π}{2}, \frac{3 π}{2}$ $s i n m x + 1 = 0 \Rightarrow s i n m x = - 1$ $0 < m x < 2 m π$ $i t m u s t h a v e t w o r o o t s \Rightarrow m = 2 ✓$
	Answered by mr W last updated on 03/Dec/23

	$\sin^{2} x - \sin m x \cos^{2} x = 1$ $\cos^{2} x (\sin m x + 1) = 0$ $\Rightarrow \cos x = 0 \Rightarrow x = \frac{π}{2}, \frac{3 π}{2}$ $\Rightarrow \sin m x + 1 = 0 \Rightarrow \sin m x = - 1 \Rightarrow m x = 2 k π - \frac{π}{2}$ $\Rightarrow x = \frac{1}{m} (2 k π - \frac{π}{2})$ $t h e r e m u s t b e t w o v a l u e s f o r x \in (0, 2 π)$ $w h i c h a r e d i f f e r e n t t h a n \frac{π}{2} a n d \frac{3 π}{2} .$ $0 < \frac{1}{m} (2 k π - \frac{π}{2}) < 2 π$ $0 < \frac{4 k - 1}{m} < 4$ $\frac{1}{m} (2 k π - \frac{π}{2}) \neq \frac{π}{2} \Rightarrow k \neq \frac{m + 1}{4}$ $\frac{1}{m} (2 k π - \frac{π}{2}) \neq \frac{3 π}{2} \Rightarrow k \neq \frac{3 m + 1}{4}$ $c a s e m > 0 :$ $0 < 4 k - 1 < 4 m$ $\frac{1}{4} < k < m + \frac{1}{4} \Rightarrow 1 ⩽ k ⩽ m$ $t w o p a i r s (k, m) :$ $m = 2 a n d k = 1, 2 .$ $o r t h r e e p a i r s (k, m) :$ $m = 3 a n d k = 1, 2, 3 (w i t h k = 1 f o r x = \frac{π}{2}) .$ $c a s e m < 0 :$ $4 m < 4 k - 1 < 0$ $m + \frac{1}{4} < k < \frac{1}{4} \Rightarrow m + 1 ⩽ k ⩽ 0$ $t w o p a i r s (k, m)$ $m = - 2 a n d k = - 1, 0 .$ $o r t h r e e p a i r s (k, m) :$ $m = - 3 a n d k = - 2, - 1, 0 (w i t h k = - 2 f o r x = \frac{π}{2})$ $s u m m a r y :$ $f o r m \in Z : m = \pm 2, \pm 3$ $f o r m \in N : m = 2, 3$
	Commented by mr W last updated on 03/Dec/23

	Commented by mr W last updated on 03/Dec/23

	Commented by mnjuly1970 last updated on 03/Dec/23

	$t h a n k s a l o t s i r$
	Commented by mnjuly1970 last updated on 03/Dec/23

	$e x c e l e n t s i r$
	Commented by MM42 last updated on 03/Dec/23

	$⋚$
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