What is the general solution for sin(x-30)cos(x-30)=(sqrt(3))/4 ?

The general solution for sin(x-30)cos(x-30)=(sqrt(3))/4 is x=60+180n,x=180n+90

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sin(x-30)cos(x-30)=(sqrt(3))/4

sin (x - 3 0^{\circ}) cos (x - 3 0^{\circ}) = \frac{3}{4}

x = 6 0^{\circ} + 18 0^{\circ} n, x = 18 0^{\circ} n + 9 0^{\circ}

Solution steps

sin (x - 3 0^{\circ}) cos (x - 3 0^{\circ}) = \frac{3}{4}

Rewrite using trig identities

\frac{sin ( 2 ( - 3 0 ^{\circ} + x ) )}{2} = \frac{3}{4}

Multiply both sides by

2

sin (2 (- 3 0^{\circ} + x)) = \frac{3}{2}

General solutions for

sin (2 (- 3 0^{\circ} + x)) = \frac{3}{2}

2 (- 3 0^{\circ} + x) = 6 0^{\circ} + 36 0^{\circ} n, 2 (- 3 0^{\circ} + x) = 12 0^{\circ} + 36 0^{\circ} n

Solve

2 (- 3 0^{\circ} + x) = 6 0^{\circ} + 36 0^{\circ} n : x = 6 0^{\circ} + 18 0^{\circ} n

Solve

2 (- 3 0^{\circ} + x) = 12 0^{\circ} + 36 0^{\circ} n : x = 18 0^{\circ} n + 9 0^{\circ}

x = 6 0^{\circ} + 18 0^{\circ} n, x = 18 0^{\circ} n + 9 0^{\circ}

cos (θ) = - \frac{7}{15}, cos (\frac{θ}{2}), 18 0^{\circ} < θ < 27 0^{\circ}

tan (x) + 2 = 0

2 6^{2} = 1 3^{2} + 1 9^{2} - (13) (19) cos (J)

4 = sec^{2} (x)

tan (\frac{x}{2}) = 4

What is the general solution for sin(x-30)cos(x-30)=(sqrt(3))/4 ?
The general solution for sin(x-30)cos(x-30)=(sqrt(3))/4 is x=60+180n,x=180n+90