Trigonometric Functions Test 30

Given that $$\sin\theta+\cos\theta=1$$

Simplifying $$\sin 3x = \cos 2x$$.

$$\sin 3x = \cos 2x$$

$$3\sin x - 4{\sin ^3}x = 1 - 2{\sin ^2}x$$

$$4{\sin ^3}x - 2{\sin ^2}x - 3\sin x + 1 = 0$$

Put $$\sin x = t$$.

$$4{t^3} - 2{t^2} - 3t + 1 = 0$$

$$\left( {t - 1} \right)\left( {4{t^2} + 2t - 1} \right) = 0$$

$$t = 1$$

Or,

$$4{t^2} + 2t - 1 = 0$$

$$t = \frac{{ - 2 \pm \sqrt {{{\left( 2 \right)}^2} - 4\left( 4 \right)\left( { - 1} \right)} }}{{2 \times 4}}$$

$$ = \frac{{ - 2 \pm \sqrt {20} }}{8}$$

$$ = \frac{{ - 2 \pm 2\sqrt 5 }}{8}$$

$$ = \frac{{ - 1 \pm \sqrt 5 }}{4}$$

Then,

$$\sin x = 1$$

$$x = \frac{\pi }{2}$$

Or,

$$\sin x = \frac{{ - 1 \pm \sqrt 5 }}{4}$$

$$\sin x = \frac{{ - 1 + \sqrt 5 }}{4}$$

$$x = \pi  + \frac{\pi }{{10}}$$

$$x = \frac{{11\pi }}{{10}}$$

Or,

$$\sin x = \frac{{ - 1 - \sqrt 5 }}{4}$$

$$x = 2\pi  - \frac{\pi }{{10}}$$

$$x = \frac{{19\pi }}{{10}}$$

So, $$x = \frac{\pi }{2}$$, $$x = \frac{{11\pi }}{{10}}$$ and $$x = \frac{{19\pi }}{{10}}$$.

Since the given interval is $$\left( {\frac{\pi }{2},\pi } \right)$$, then, $$x = \frac{\pi }{2}$$ is the only solution.

Therefore, the number of solutions of $$\sin 3x = \cos 2x$$ is 1.

$$\cfrac { \cos A }{ 1-\sin A } =\cfrac { \cos A(1+\sin A) }{ (1-\sin A)(1+\sin A) } =\cfrac { \cos A(1+\sin A) }{ { \cos }^{ 2 }A } \\ \cfrac { 1+\sin A }{ \cos A } =\cfrac { ({ \sin }^{ 2 }\cfrac { A }{ 2 } +{ \cos }^{ 2 }\cfrac { A }{ 2 } +2\sin\cfrac { A }{ 2 } .\cos\cfrac { A }{ 2 } ) }{ { \cos }^{ 2 }\cfrac { A }{ 2 } -{ \sin }^{ 2 }\cfrac { A }{ 2 }  } \\ \cfrac { { (\sin\cfrac { A }{ 2 } +\cos\cfrac { A }{ 2 } ) }^{ 2 } }{ \left( \cos\cfrac { A }{ 2 } +\sin\cfrac { A }{ 2 }  \right) \left( \cos\cfrac { A }{ 2 } +\sin\cfrac { A }{ 2 }  \right)  } \\ \\ $$