# Find all values of x in the interval [0,2pi] that satisfy the equation cos2x cos3x = sin2x sin3x (exact solutions required)

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### 3 Answers

Sorry for the slip above.

Take note of this correction.

Cosine function is zero at angles `pi/2` and `(3pi)/2` . So,

`5x = pi/2, (3pi)/2, (5pi)/2, (7pi)/2, (9pi)/2, ``...``((2k+1)pi)/2` (where k is any integer)

`x=pi/10, (3pi)/10, (5pi)/10, (7pi)/10, (9pi)/10,` `...` `((2k+1)pi)/10`

Since angle x must be in the interval [0, 2pi], the possible values of k in ((2k+1)pi/10 must be solved.

To do so, set ((2k+1)pi/10 greater than or equal to zero to get the possible minimum value of k

`((2k+1)pi)/10gt=0`

`2k+1gt=0`

`2kgt=-1`

`kgt=-1/2`

Since k is an integer, the minimum value is 0.

Also,set `((2k+1)pi)/10` less than or equal to 2pi to determine the maximum value of k.

`((2k+1)pi)/10<=2pi`

`2k+1<=20`

`2k<=19`

`k<=19/2`

Since k is an integer, then its maximum value is 9. So, the interval for k is `0lt=klt=9` .

**Hence, the solutions to the given equation are `x={pi/10, (3pi)/10, (5pi)/10, (7pi)/10, (9pi)/10, (11pi)/10, (13pi)/10, (15pi)/10, (17pi)/10, (19pi)/10}` .**

`cos(2x) cos(3x) = sin(2x) sin(3x)`

To solve. set one side equal to zero.

`cos(2x) cos(3x)-sin(2x) sin(3x) =0`

Then, apply the identity for sum of two angles of cosine which is `cos(A+B)=cosAcosB - sinAsinB` .

`cos(2x+3x)=0`

`cos(5x)=0`

Since cosine is zero at 0, pi and 2pi, then:

`5x=0, pi, 2pi, 3pi, 4pi, 5pi ... pik` (where k is any integer)

`x=0,pi/5,(2pi)/5, (3pi)/5, (4pi)/5, pi ... pi/5k`

Since x must be in the interval `[0,2pi]` , set `pi/5k` to the boundary of the interval to get the range of k.

`pi/5k=0` and `pi/5k=2pi`

`k=0 ` `k=10`

So, the interval for k is `0lt=klt=10` .

**Hence, `x = {0,pi/5,(2pi)/5, (3pi)/5, (4pi)/5, pi , (6pi)/5, (7pi)/5, (8pi)/5, (9pi)/5, 2pi}` .**

`cos2xcos3x=sin2xsin3x`

`cos2xcos3x-sin2xsin3x=0`

`cos5x=0` `5x=pi/2` `5x=3/2pi`

`x=3/10 pi` `x=1/10 pi`

`x=1/10 pi +kpi/5`