What is the general solution for tan(2x-5)=sin^2(30)+sin^2(60) ?

The general solution for tan(2x-5)=sin^2(30)+sin^2(60) is x= 5/2+22.5+(180n)/2

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tan(2x-5)=sin^2(30)+sin^2(60)

tan (2 x - 5) = sin^{2} (3 0^{\circ}) + sin^{2} (6 0^{\circ})

x = \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

Solution steps

tan (2 x - 5) = sin^{2} (3 0^{\circ}) + sin^{2} (6 0^{\circ})

sin (3 0^{\circ}) = \frac{1}{2}

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

tan (2 x - 5) = (\frac{1}{2})^{2} + (\frac{3}{2})^{2}

Simplify

(\frac{1}{2})^{2} + (\frac{3}{2})^{2} : 1

General solutions for

tan (2 x - 5) = 1

2 x - 5 = 4 5^{\circ} + 18 0^{\circ} n

Solve

2 x - 5 = 4 5^{\circ} + 18 0^{\circ} n : x = \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

5 sin (x) (sin (x) - 1) = 0

5 tan^{2} (x) - 9 sec (x) + 2 = 0

- sin (A) + 2 cos (A) sin (A) = 0

sin (2 θ) = - \frac{1}{2}, \forall 0 \leq θ < 2 π

t, 2 - 2 cosh (t - 2) = 0

What is the general solution for tan(2x-5)=sin^2(30)+sin^2(60) ?
The general solution for tan(2x-5)=sin^2(30)+sin^2(60) is x= 5/2+22.5+(180n)/2