solve for theda given 0<theda<2pi
cos theta= - (sqrt 3)/ 2
you know that cos π/6 = √3/2
You also know that cos(x) < 0 in QII,QIII
So, ...
so ould it be sin pi.6 and tan pi/6??
To solve for θ in the equation cos θ = -√3/2, where 0 < θ < 2π, you can use the inverse cosine function (also known as arccosine or cos^-1).
Step 1: Take the inverse cosine of both sides of the equation:
θ = arccos(-√3/2)
Step 2: Evaluate the inverse cosine on the right side using a calculator or mathematical table:
θ ≈ 5π/3 (approximately 2.0944)
Step 3: Check if the solution is within the given range 0 < θ < 2π:
The solution, 5π/3, is within the given range since 0 < 5π/3 < 2π.
Therefore, the value of θ that satisfies the equation cos θ = -√3/2, where 0 < θ < 2π, is approximately 5π/3.