4cos^2(x)+cos(2x)-7cos(x)=-2,0<= x<= 360

Solution

4 cos^{2} (x) + cos (2 x) - 7 cos (x) = - 2, 0^{\circ} \leq x \leq 36 0^{\circ}

x = 0, x = 36 0^{\circ}, x = 1.40334 \dots, x = 36 0^{\circ} - 1.40334 \dots

Solution steps

4 cos^{2} (x) + cos (2 x) - 7 cos (x) = - 2, 0^{\circ} \leq x \leq 36 0^{\circ}

Subtract

- 2

from both sides

4 cos^{2} (x) + cos (2 x) - 7 cos (x) + 2 = 0

Rewrite using trig identities

1 + 6 cos^{2} (x) - 7 cos (x) = 0

Solve by substitution

cos (x) = 1, cos (x) = \frac{1}{6}

cos (x) = 1, 0 \leq x \leq 36 0^{\circ} : x = 0, x = 36 0^{\circ}

cos (x) = \frac{1}{6}, 0 \leq x \leq 36 0^{\circ} : x = arccos (\frac{1}{6}), x = 36 0^{\circ} - arccos (\frac{1}{6})

Combine all the solutions

x = 0, x = 36 0^{\circ}, x = arccos (\frac{1}{6}), x = 36 0^{\circ} - arccos (\frac{1}{6})

Show solutions in decimal form

x = 0, x = 36 0^{\circ}, x = 1.40334 \dots, x = 36 0^{\circ} - 1.40334 \dots

\frac{16}{sin ( a )} = \frac{12}{sin ( 3 1 ^{\circ} )}

tan (x) = \frac{25}{10}

sin (x) - 1 = 0, 0 \leq x < 2 π

3 cos (θ) = 2 cos (θ)

\frac{26.4575}{sin ( 6 0 ^{\circ} )} = \frac{20}{sin ( B )}