What is the general solution for 4cos(2θ)-10cos(θ)+14=7,0<= θ<360 ?

The general solution for 4cos(2θ)-10cos(θ)+14=7,0<= θ<360 is θ=0.72273…,θ=360-0.72273…,θ=60,θ=300

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4cos(2θ)-10cos(θ)+14=7,0<= θ<360

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots, θ = 6 0^{\circ}, θ = 30 0^{\circ}

Solution steps

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

Subtract

7

from both sides

4 cos (2 θ) - 10 cos (θ) + 7 = 0

Rewrite using trig identities

3 - 10 cos (θ) + 8 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = \frac{3}{4}, cos (θ) = \frac{1}{2}

cos (θ) = \frac{3}{4}, 0 \leq θ < 36 0^{\circ} : θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4})

cos (θ) = \frac{1}{2}, 0 \leq θ < 36 0^{\circ} : θ = 6 0^{\circ}, θ = 30 0^{\circ}

Combine all the solutions

θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4}), θ = 6 0^{\circ}, θ = 30 0^{\circ}

Show solutions in decimal form

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots, θ = 6 0^{\circ}, θ = 30 0^{\circ}

x, Z = sin (x y^{2})

2 tan^{2} (θ) = 2

0 = (6 cos (30 t) + 2 sin (30 t))

- 2 sin (θ) = 2

tan (2 x - \frac{π}{6}) = \frac{1}{3 ^{\frac{1}{2}}}

What is the general solution for 4cos(2θ)-10cos(θ)+14=7,0<= θ<360 ?
The general solution for 4cos(2θ)-10cos(θ)+14=7,0<= θ<360 is θ=0.72273…,θ=360-0.72273…,θ=60,θ=300