What is the general solution for sec(2x-10)=csc(50) ?

The general solution for sec(2x-10)=csc(50) is x=180n+5+(0.69813…)/2 ,x=180+180n+5-(0.69813…)/2

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sec(2x-10)=csc(50)

sec (2 x - 1 0^{\circ}) = csc (5 0^{\circ})

x = 18 0^{\circ} n + 5^{\circ} + \frac{0.69813 \dots}{2}, x = 18 0^{\circ} + 18 0^{\circ} n + 5^{\circ} - \frac{0.69813 \dots}{2}

Solution steps

sec (2 x - 1 0^{\circ}) = csc (5 0^{\circ})

Apply trig inverse properties

2 x - 1 0^{\circ} = arcsec (csc (5 0^{\circ})) + 36 0^{\circ} n, 2 x - 1 0^{\circ} = 36 0^{\circ} - arcsec (csc (5 0^{\circ})) + 36 0^{\circ} n

Solve

2 x - 1 0^{\circ} = arcsec (csc (5 0^{\circ})) + 36 0^{\circ} n : x = 18 0^{\circ} n + 5^{\circ} + \frac{arcsec ( csc ( 5 0 ^{\circ} ) )}{2}

Solve

2 x - 1 0^{\circ} = 36 0^{\circ} - arcsec (csc (5 0^{\circ})) + 36 0^{\circ} n : x = 18 0^{\circ} + 18 0^{\circ} n + 5^{\circ} - \frac{arcsec ( csc ( 5 0 ^{\circ} ) )}{2}

x = 18 0^{\circ} n + 5^{\circ} + \frac{arcsec ( csc ( 5 0 ^{\circ} ) )}{2}, x = 18 0^{\circ} + 18 0^{\circ} n + 5^{\circ} - \frac{arcsec ( csc ( 5 0 ^{\circ} ) )}{2}

Show solutions in decimal form

x = 18 0^{\circ} n + 5^{\circ} + \frac{0.69813 \dots}{2}, x = 18 0^{\circ} + 18 0^{\circ} n + 5^{\circ} - \frac{0.69813 \dots}{2}

7 sec (x) - 7 = 0, 0 \leq x \leq 2 π

cos (x) + cos (x) = cos (2 x)

csc^{2} (2 x - 0.6) = 16

tan (x - 1) = 2

5 sin^{2} (x) + 10 sin (x) + 2 = 0

What is the general solution for sec(2x-10)=csc(50) ?
The general solution for sec(2x-10)=csc(50) is x=180n+5+(0.69813…)/2 ,x=180+180n+5-(0.69813…)/2