That's a slightly odd looking equation. It's not clear if the angles are degrees or radians. In particular the -1 and the 7 need their units clarified. The usual convention is unitless means radians, but you don't usually see 1 radian and 7 radians being tossed around with no pis. I'm going with degrees.
Solve sin(4x-1^circ)=cos(2x + 7^circ)
What I always remember is cos x = cos x has solutions x = \pm a + 360 ^circ k quad for integer k.
We use complementary angles to turn the sine to a cosine:
cos(90^circ - (4x - 1^circ)) = cos (2x + 7^circ)
Now we apply our solution:
90^circ - (4x - 1^circ) = \pm (2x + 7^circ) + 360^circ k
It's simpler just to handle + and - separately. Plus first:
90^circ - (4x - 1^circ) = (2x + 7^circ) + 360^circ k
90^circ - (4x - 1^circ) = (2x + 7^circ) + 360^circ k
-4x - 2x =-90^circ - 1^ circ + 7^circ + 360^circ k
-6x = -84^circ + 360^circ k
x= 14 ^circ + 60^circ k
k ranges over the integers so it's ok how I flipped its sign to keep the plus sign.
Now the - part of the pm:
90^circ - (4x - 1^circ) = - (2x + 7^circ) + 360^circ k
-2x = - 98^circ + 360^circ k
x = 49^circ + 180^circ k
The full solution to sin(4x-1^circ)=cos(2x + 7^circ) is
x= 14 ^circ + 60^circ k or x = 49^circ + 180^circ k quad for integer k.
Check:
sin(4(14+60k)-1) = sin(55-240k) = cos(90-55-240k) = cos(35-240k)
cos(2(14 + 60k) + 7) = cos(35 + 120k) quad sqrt
Those are identical for a given k.
sin(4(49+180k)-1) = sin(195) = cos(90-195) = cos(105)
cos(2(49+180k)+7)=cos(105) quad sqrt
