2cos(2x)+sqrt(2)=0,0<= x<= 2pi

Solution

2 cos (2 x) + 2 = 0, 0 \leq x \leq 2 π

x = \frac{3 π}{8}, x = \frac{5 π}{8}, x = \frac{11 π}{8}, x = \frac{13 π}{8}

Solution steps

2 cos (2 x) + 2 = 0, 0 \leq x \leq 2 π

Move

2

to the right side

2 cos (2 x) = - 2

Divide both sides by

2

cos (2 x) = - \frac{2}{2}

General solutions for

cos (2 x) = - \frac{2}{2}

2 x = \frac{3 π}{4} + 2 πn, 2 x = \frac{5 π}{4} + 2 πn

Solve

2 x = \frac{3 π}{4} + 2 πn : x = \frac{3 π}{8} + πn

Solve

2 x = \frac{5 π}{4} + 2 πn : x = \frac{5 π}{8} + πn

x = \frac{3 π}{8} + πn, x = \frac{5 π}{8} + πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{8}, x = \frac{5 π}{8}, x = \frac{11 π}{8}, x = \frac{13 π}{8}

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

\frac{tan ( x )}{sin ( x )} = cos (x)

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

tan (θ) = - 1 - 3

y, v = (1 - 4 cos (5 y))^{- \frac{1}{2}}