Find all solutions of the equation in the interval [0,2pi) The equation is 2cos(theta)+sqr root of 2 =0

Answer 1

cos(theta) = - sqrt(2)/2

Draw a figure...a right triangle with hypotenuse = 2 and side opposite theta = sqrt(2).

What is the other side?

What triangle is this?

Now where is the cos negative? The mnemonic is "all sin tan cos".

Now you should know your answers.