×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
The equation 4 ( sin x cos 3 x – cos x sin 3 x + 1) = 3 has the general solution: (n pi / 4) + ( 8 /pi) ( n pi/ 2) + ( 3 pi /8) n pi (+/– ) (5 pi / 8) (n pi/4) + ( 7 pi / 4)
The equation 4 ( sin x cos3x – cos x sin3x + 1) = 3 has the general solution:	(n pi / 4) + ( 8 /pi)	( n  pi/ 2) + ( 3 pi /8)	 n pi (+/– ) (5 pi / 8)	(n pi/4) + ( 7 pi / 4)

```
2 years ago

Arun
25768 Points
```							Dear student 4 sin x cos x (cos^2 x – sin^2 x) = 3 – 4 2 sin 2x cos 2x = – 1 sin 4x = – 1 hence 4x = 2n * pi + 3pi/2 x = n * pi / 2 + 3 * pi /8
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Trigonometry

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions