# Thread: trigo problem no. 21

1. ## trigo problem no. 21

p195 q27 ex9a

i've got problem again
p.s. or is there any other method besides my method 1 and 2 ?

2. Originally Posted by afeasfaerw23231233
p195 q27 ex9a

i've got problem again
p.s. or is there any other method besides my method 1 and 2 ?
I assume 0< x < 360 ......

In method 1, you've neglected the solutions corresponding to 5x/2 = 360, 540 and 700 ....

In method 2, if you don't use the calculator then you just have to know the special angles that those exact surd values correspond to ......

3. Originally Posted by afeasfaerw23231233
p195 q27 ex9a

i've got problem again
p.s. or is there any other method besides my method 1 and 2 ?
Method 3 (my favourite):

sin(2x) = -sin(3x) = sin(-3x).

Therefore (working in degrees):

Case 1:

2x = -3x + 360n, n is an integer

=> 5x = 360n

=> x = 72n

=> x = 0, 72, 144, 216, 288 (assuming 0 < x < 360).

Case 2:

2x = 180 - (-3x) + 360n, n is an integer

=> -x = 180 + 360n = 180(1 + 2n)

=> x = -180(2n + 1)

=> x = 180 (assuming 0 < x< 360).