What is the general solution for 2*sin(2x+15)=1 ?

The general solution for 2*sin(2x+15)=1 is x=180n+7.5,x=180n+67.5

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2*sin(2x+15)=1

2 \cdot sin (2 x + 1 5^{\circ}) = 1

x = 18 0^{\circ} n + 7. 5^{\circ}, x = 18 0^{\circ} n + 67. 5^{\circ}

Solution steps

2 sin (2 x + 1 5^{\circ}) = 1

Divide both sides by

2

sin (2 x + 1 5^{\circ}) = \frac{1}{2}

General solutions for

sin (2 x + 1 5^{\circ}) = \frac{1}{2}

2 x + 1 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, 2 x + 1 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

Solve

2 x + 1 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n : x = 18 0^{\circ} n + 7. 5^{\circ}

Solve

2 x + 1 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n : x = 18 0^{\circ} n + 67. 5^{\circ}

x = 18 0^{\circ} n + 7. 5^{\circ}, x = 18 0^{\circ} n + 67. 5^{\circ}

cos (2 x) + 3 cos (x) + 2 = 0, \frac{π}{2} \leq x \leq π

2 cos (a) = 3

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

sin^{2} (4 x) = 1

sinh (x) = \frac{2}{3}

What is the general solution for 2*sin(2x+15)=1 ?
The general solution for 2*sin(2x+15)=1 is x=180n+7.5,x=180n+67.5