What is the general solution for tan(a+5)=sqrt(2sin(30)+sec(245)) ?

The general solution for tan(a+5)=sqrt(2sin(30)+sec(245)) is No Solution for a\in\mathbb{R}

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tan(a+5)=sqrt(2sin(30)+sec(245))

tan (a + 5^{\circ}) = 2 sin (3 0^{\circ}) + sec (24 5^{\circ})

No Solution for a \in R

Solution steps

tan (a + 5^{\circ}) = 2 sin (3 0^{\circ}) + sec (24 5^{\circ})

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

sin (3 0^{\circ})

Use the following trivial identity:

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

sin (3 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{1}{2}

tan (a + 5^{\circ}) = 2 \cdot \frac{1}{2} + sec (24 5^{\circ})

Simplify

2 \cdot \frac{1}{2} + sec (24 5^{\circ}) : i - 1 - sec (24 5^{\circ})

2 \cdot \frac{1}{2} + sec (24 5^{\circ})

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 1 + sec (24 5^{\circ})

Apply imaginary number rule:

- a = i a

= i - sec (24 5^{\circ}) - 1

No Solution for a \in R

cot (x) = tan (2 5^{\circ})

2 sin (2 θ) = cos (2 θ + 3 0^{\circ})

cos^{2} (x) = \frac{3}{2}

2 sin^{2} (x) + cos^{2} (x) + cos (x) - 1 = 0

14 4^{2} = 43.2 7^{2} + 7 2^{2} - 2 \cdot 43.27 \cdot 72 \cdot cos (x)

What is the general solution for tan(a+5)=sqrt(2sin(30)+sec(245)) ?
The general solution for tan(a+5)=sqrt(2sin(30)+sec(245)) is No Solution for a\in\mathbb{R}