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 #pi#s. 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#
