What is the general solution for sec(2θ)=csc(70) ?

The general solution for sec(2θ)=csc(70) is θ=(0.34906…)/2+180n,θ=180-(0.34906…)/2+180n

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sec(2θ)=csc(70)

sec (2 θ) = csc (7 0^{\circ})

θ = \frac{0.34906 \dots}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{0.34906 \dots}{2} + 18 0^{\circ} n

Solution steps

sec (2 θ) = csc (7 0^{\circ})

Apply trig inverse properties

2 θ = arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n, 2 θ = 36 0^{\circ} - arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n

Solve

2 θ = arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n : θ = \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

Solve

2 θ = 36 0^{\circ} - arcsec (csc (7 0^{\circ})) + 36 0^{\circ} n : θ = 18 0^{\circ} - \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

θ = \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{arcsec ( csc ( 7 0 ^{\circ} ) )}{2} + 18 0^{\circ} n

Show solutions in decimal form

θ = \frac{0.34906 \dots}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{0.34906 \dots}{2} + 18 0^{\circ} n

- 4 cos^{2} (θ) + 1 = 0

sin (2 x) - sin (4 x) + sin (6 x) = 0

2 csc (x) - 3 = 0

cos (θ) = \frac{8}{12}

sin (x) sin (2 x) = 0

What is the general solution for sec(2θ)=csc(70) ?
The general solution for sec(2θ)=csc(70) is θ=(0.34906…)/2+180n,θ=180-(0.34906…)/2+180n