What is the general solution for 6sin(2x)=3cos(30) ?

The general solution for 6sin(2x)=3cos(30) is x=(0.44783…)/2+180n,x=90-(0.44783…)/2+180n

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6sin(2x)=3cos(30)

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

x = \frac{0.44783 \dots}{2} + 18 0^{\circ} n, x = 9 0^{\circ} - \frac{0.44783 \dots}{2} + 18 0^{\circ} n

Solution steps

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

cos (3 0^{\circ}) = \frac{3}{2}

6 sin (2 x) = 3 \cdot \frac{3}{2}

Divide both sides by

6

sin (2 x) = \frac{3}{4}

Apply trig inverse properties

2 x = arcsin (\frac{3}{4}) + 36 0^{\circ} n, 2 x = 18 0^{\circ} - arcsin (\frac{3}{4}) + 36 0^{\circ} n

Solve

2 x = arcsin (\frac{3}{4}) + 36 0^{\circ} n : x = \frac{arcsin ( \frac{3}{4} )}{2} + 18 0^{\circ} n

Solve

2 x = 18 0^{\circ} - arcsin (\frac{3}{4}) + 36 0^{\circ} n : x = 9 0^{\circ} - \frac{arcsin ( \frac{3}{4} )}{2} + 18 0^{\circ} n

x = \frac{arcsin ( \frac{3}{4} )}{2} + 18 0^{\circ} n, x = 9 0^{\circ} - \frac{arcsin ( \frac{3}{4} )}{2} + 18 0^{\circ} n

Show solutions in decimal form

x = \frac{0.44783 \dots}{2} + 18 0^{\circ} n, x = 9 0^{\circ} - \frac{0.44783 \dots}{2} + 18 0^{\circ} n

cos (y) = - 1

cot (x) = - \frac{2}{3}

tan (\frac{3 θ}{4}) = \frac{- 3}{3}

30 = 15 cos (\frac{2 π}{365} t) + 43

sin (x) = + 1 e

What is the general solution for 6sin(2x)=3cos(30) ?
The general solution for 6sin(2x)=3cos(30) is x=(0.44783…)/2+180n,x=90-(0.44783…)/2+180n