(2sin(x)+5)(3cos(x)+3)(tan^2(x)-2)=0

Solution

(2 sin (x) + 5) (3 cos (x) + 3) (tan^{2} (x) - 2) = 0

x = π + 2 πn, x = 0.95531 \dots + πn, x = - 0.95531 \dots + πn

Solution steps

(2 sin (x) + 5) (3 cos (x) + 3) (tan^{2} (x) - 2) = 0

Solving each part separately

2 sin (x) + 5 = 0 or 3 cos (x) + 3 = 0 or tan^{2} (x) - 2 = 0

2 sin (x) + 5 = 0 :

No Solution

3 cos (x) + 3 = 0 : x = π + 2 πn

tan^{2} (x) - 2 = 0 : x = arctan (2) + πn, x = arctan (- 2) + πn

Combine all the solutions

x = π + 2 πn, x = arctan (2) + πn, x = arctan (- 2) + πn

Show solutions in decimal form

x = π + 2 πn, x = 0.95531 \dots + πn, x = - 0.95531 \dots + πn

3 sin (x) + 3 cos (2 x) = 2

tan (3 x + 2 5^{\circ}) \cdot 16 0^{\circ} - 1 = 0

5 cos (x) = 1

cos (θ) = \frac{2}{11}

3 cos (x) + 5 = 7, 0 \leq x \leq 2 π