How do you solve #2cos^2(2x)+cos(2x)-1=0 #?

Question

Dylan Arnold · Accepted Answer

#x = pi/6 + kpi or pi/3 + kpi or pi/2 + kpi#,

for any integer #k#.

Let #cos(2x)# be #y#. So the equation becomes a standard quadratic #2y^2 + y - 1 = 0# Factorize to find the zeroes. #(2y - 1) * (y + 1) = 0# This means that #y = 1/2# or #y = -1#. Now change back #y# to #cos(2x)#. So we get #cos(2x) = 1/2# or #cos(2x) = -1# Next we find the basic angles. #cos^{-1}(1/2) = pi/3# #cos^{-1}(-1) = pi# With that, we solve #cos(2x) = 1/2# first. #2x = pi/3 + 2kpi or {2pi}/3 + 2kpi#, for any integer #k#. #x = pi/6 + kpi or pi/3 + kpi# Now, we solve #cos(2x) = -1#. #2x = pi + 2kpi#, for any integer #k#. #x = pi/2 + kpi# Combining the solution gives #x = pi/6 + kpi or pi/3 + kpi or pi/2 + kpi#, for any integer #k#.

Victoria Adebayo · Answer

undefined To solve $2\cos^2(2x) + \cos(2x) - 1 = 0$, you can treat it as a quadratic equation in terms of $\cos(2x)$.

1. Let $u = \cos(2x)$.
2. Rewrite the equation in terms of $u$:
$$2u^2 + u - 1 = 0$$
3. Solve this quadratic equation for $u$.
4. Once you find the solutions for $u$, substitute them back into $u = \cos(2x)$ and solve for $x$.

This will give you the solutions for $x$.