Question

Solve this equation using sum to product formulas: cos3x + sin2x - sin6x + cos5x = 0

Answer 1

Use formula:

$\cos(c) + \cos(d) = 2 \cos( \frac{c + d}{2} ) . \cos( \frac{c - d}{2} )$

$\sin(c) - \sin(d) = 2 \cos( \frac{c + d}{2} ) . \sin( \frac{c - d}{2} )$

then solve equation and you will get simpler terms in sin and cos

Like,

cos4x(sinx-cosx)=0

Solve

cos4x=0

then sinx-cosx=0

which are easier to solve.

Answer 2

Answer:

x=(nπ/2)±15

x=(2nπ+90)/3

x=90-2nπ

Steps:

cos3x+sin2x-sin6x+cos5x=0

(cos3x+cos5x)+(sin2x-sin6x)=0

equation(1)

Use formula:

$\cos(c) + \cos(d) = 2 \cos( \frac{c + d}{2} ) . \cos( \frac{c - d}{2} )$

$\sin(c) - \sin(d) = 2 \cos( \frac{c + d}{2} ) . \sin( \frac{c - d}{2} )$

So in equation (1),if c=3x ,d=5x,C=2x,D=6x

》2cos{(c+d)/2}.cos{(c-d)/2}+2cos{(C+D)/2}.sin{(C-D)/2}=0

》2cos{(3x+5x)/2}.cos{(3x-5x)/2}+2cos{(2x+6x)/2}.sin{(2x-6x)/2}=0

》2cos(4x).cos(-x)+2cos(4x).sin(-2x)=0

》2cos(4x)[cos(-x)+sin(-2x)]=0

》2cos(4x)[cos(x)-sin(2x)]=0

1) Either:

2cos(4x)=0

cos(4x)=0/2

cos(4x)=0

cos(4x)=cos(90)

General solution for such case is:X=2n±a

So,

4x=2nπ±90

x=(2nπ±90)/4

x=(nπ/2)±15

2) Or:

cos(x)-sin(2x)=0

cos(x)=sin(2x)

General solution for such case is:X=2n±a

So,

x=2nπ±(90-2x)

x=2nπ±90±2x

x±2x=2nπ±90

Take +ve sign,

x+2x=2nπ+90

3x=2nπ+90

x=(2nπ+90)/3

Take -ve sign,

x-2x=2nπ-90

-x=2nπ-90

x=(2nπ-90)/(-1)

x=90-2nπ