What is the general solution for sin(2x+7)=cos(4x-7) ?

The general solution for sin(2x+7)=cos(4x-7) is x=(4pin+pi}{12},x=-\frac{pi+4pin-28)/4

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sin(2x+7)=cos(4x-7)

sin (2 x + 7) = cos (4 x - 7)

x = \frac{4 πn + π}{12}, x = - \frac{π + 4 πn - 28}{4}

Solution steps

sin (2 x + 7) = cos (4 x - 7)

Rewrite using trig identities

sin (2 x + 7) = sin (\frac{π}{2} - (4 x - 7))

Apply trig inverse properties

2 x + 7 = \frac{π}{2} - (4 x - 7) + 2 πn, 2 x + 7 = π - (\frac{π}{2} - (4 x - 7)) + 2 πn

2 x + 7 = \frac{π}{2} - (4 x - 7) + 2 πn : x = \frac{4 πn + π}{12}

2 x + 7 = π - (\frac{π}{2} - (4 x - 7)) + 2 πn : x = - \frac{π + 4 πn - 28}{4}

x = \frac{4 πn + π}{12}, x = - \frac{π + 4 πn - 28}{4}

x, cot (x) = 2

- 2 sin^{2} (x) + sin (x) = 0

2 sin (x + 6 0^{\circ}) = cos (x + 3 0^{\circ})

cot^{2} (θ) - 3 csc (θ) = - 3

tan^{3} (x) = \frac{1}{3} tan (x)

What is the general solution for sin(2x+7)=cos(4x-7) ?
The general solution for sin(2x+7)=cos(4x-7) is x=(4pin+pi}{12},x=-\frac{pi+4pin-28)/4