1

### JEE Main 2018 (Online) 15th April Evening Slot

The number of solutions of sin3x = cos 2x, in the interval $\left( {{\pi \over 2},\pi } \right)$ is :
A
1
B
2
C
3
D
4

## Explanation

sin 3x = cos 2x

$\Rightarrow $$\,\,\, 3 sin x - 4 sin3 x = 1 - 2 sin2 x \Rightarrow$$\,\,\,$ 4 sin3 x $-$ 2 sin2 x $-$ 3 sin x + 1 = 0

$\Rightarrow $$\,\,\, sin x = 1, {{ - 2 \pm 2\sqrt 5 } \over 8} In the interval \left( {{\pi \over 2},\pi } \right), sin x = {{ - 2 + 2\sqrt 5 } \over 8} So, there is only one solution. 2 MCQ (Single Correct Answer) ### JEE Main 2019 (Online) 9th January Morning Slot For any \theta \in \left( {{\pi \over 4},{\pi \over 2}} \right), the expression 3{(\cos \theta - \sin \theta )^4}$$ + 6{(\sin \theta + \cos \theta )^2} + 4{\sin ^6}\theta$

equals :
A
13 – 4 cos2$\theta$ + 6sin2$\theta$cos2$\theta$
B
13 – 4 cos6$\theta$
C
13 – 4 cos2$\theta$ + 6cos2$\theta$
D
13 – 4 cos4$\theta$ + 2sin2$\theta$cos2$\theta$

## Explanation

Given,

3(sin$\theta$ $-$ cos$\theta$)4 + 6(sin$\theta$ + cos$\theta$)2 + 4sin6$\theta$

= 3[(sin$\theta$ $-$ cos$\theta$)2]2 + 6 (sin2$\theta$ + cos2$\theta$ + 2sin$\theta$cos$\theta$) + 4sin6$\theta$

= 3[sin2$\theta$ + cos2$\theta$ $-$2sin$\theta$cos$\theta$]2 + 6(1 + sin2$\theta$) + 4sin6$\theta$

= 3(1 $-$ sin2$\theta$)2 + 6(1 + sin2$\theta$) + 4sin6$\theta$

= 3 (1 $-$ 2 sin2$\theta$ + sin22$\theta$) + 6 + 6sin2$\theta$ + 4sin6$\theta$

= 3 $-$ 6sin2$\theta$ + 3sin22$\theta$ + 6 + 6sin2$\theta$ + 4sin6$\theta$

= 9 + 3sin22$\theta$ + 4 sin6$\theta$

= 9 + 3(2sin$\theta$cos$\theta$)2 + 4(1 $-$ cos2$\theta$)3

= 9 + 12sin2$\theta$ cos2$\theta$ + 4 (1 $-$ cos6$\theta$ $-$ 3cos2$\theta$ + 3cos4$\theta$)

= 13 + 12 (1 $-$ cos2$\theta$ $-$ 4cos6$\theta$ $-$ 12cos$\theta$ + 12 cos4$\theta$

= 13 + 12 cos2$\theta$ $-$ 12 cos4$\theta$ $-$ 4cos6$\theta$ $-$ 12 cos2$\theta$ + 12 cos4$\theta$

= 13 $-$ 4 cos6$\theta$
3

### JEE Main 2019 (Online) 9th January Evening Slot

If  0 $\le$ x < ${\pi \over 2}$,  then the number of values of x for which sin x $-$ sin 2x + sin 3x = 0, is :
A
3
B
1
C
4
D
2

## Explanation

sin x $-$ sin 2x + sin 3x = 0        $x \in \left[ {0,{\pi \over 2}} \right)$

$\Rightarrow$  (sin3x + sinx) $-$ sin2x = 0

$\Rightarrow$  2sin2x.cos2x $-$ sin2x = 0

$\Rightarrow$  sin2x (2cosx $-$ 1) = 0

sin 2x = 0

x = 0

and cos x = ${1 \over 2}$

and x = ${\pi \over 3}$

two solutions
4

The sum of all values of $\theta$ $\in $$\left( {0,{\pi \over 2}} \right) satisfying sin2 2\theta + cos4 2\theta = {3 \over 4} is - A {{5\pi } \over 4} B {\pi \over 2} C \pi D {{3\pi } \over 8} ## Explanation sin22\theta + cos42\theta = {3 \over 4},$$\theta$ $\in$ $\left( {0,{\pi \over 2}} \right)$

$\Rightarrow$  1 $-$ cos22$\theta$ + cos42$\theta$ = ${3 \over 4}$

$\Rightarrow$  4cos2$\theta$ $-$ 4cos22$\theta$ + 1 = 0

$\Rightarrow$  (2cos22$\theta$ $-$ 1)2 = 0

$\Rightarrow$  cos22$\theta$ = ${1 \over 2}$ = cos2${{\pi \over 4}}$

$\Rightarrow$  2$\theta$ = n$\pi$ $\pm$ ${\pi \over 4}$, n $\in$ I

$\Rightarrow$  $\theta$ = ${{n\pi } \over 2} \pm {\pi \over 8}$

$\Rightarrow$  $\theta$ = ${\pi \over 8},{\pi \over 2} - {\pi \over 8}$

Sum of solutions ${\pi \over 2}$