What is the general solution for solvefor θ,cos(θ/2+30)=sin(θ) ?

The general solution for solvefor θ,cos(θ/2+30)=sin(θ) is θ=(360+2160n)/9 ,θ=(720+2160n)/3

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solvefor θ,cos(θ/2+30)=sin(θ)

solve for

θ, cos (\frac{θ}{2} + 3 0^{\circ}) = sin (θ)

θ = \frac{36 0 ^{\circ} + 216 0 ^{\circ} n}{9}, θ = \frac{72 0 ^{\circ} + 216 0 ^{\circ} n}{3}

Solution steps

cos (\frac{θ}{2} + 3 0^{\circ}) = sin (θ)

Rewrite using trig identities

cos (\frac{θ}{2} + 3 0^{\circ}) = sin (9 0^{\circ} - (\frac{θ}{2} + 3 0^{\circ}))

Apply trig inverse properties

θ = 9 0^{\circ} - (\frac{θ}{2} + 3 0^{\circ}) + 36 0^{\circ} n, θ = 18 0^{\circ} - (9 0^{\circ} - (\frac{θ}{2} + 3 0^{\circ})) + 36 0^{\circ} n

θ = 9 0^{\circ} - (\frac{θ}{2} + 3 0^{\circ}) + 36 0^{\circ} n : θ = \frac{36 0 ^{\circ} + 216 0 ^{\circ} n}{9}

θ = 18 0^{\circ} - (9 0^{\circ} - (\frac{θ}{2} + 3 0^{\circ})) + 36 0^{\circ} n : θ = \frac{72 0 ^{\circ} + 216 0 ^{\circ} n}{3}

θ = \frac{36 0 ^{\circ} + 216 0 ^{\circ} n}{9}, θ = \frac{72 0 ^{\circ} + 216 0 ^{\circ} n}{3}

5 sin^{2} (θ) = 1 + 3 sin^{2} (θ)

5 sin (2 θ) - 8 sin (θ) = 0

tan (θ) = \frac{5 3}{5}

f, r = \frac{8}{sec ( f ^{0} )}

2 sin (x) - 2 = 0, 0 \leq x \leq 2 π

What is the general solution for solvefor θ,cos(θ/2+30)=sin(θ) ?
The general solution for solvefor θ,cos(θ/2+30)=sin(θ) is θ=(360+2160n)/9 ,θ=(720+2160n)/3