How do you solve #2cos^2(x)+ 4sin^2(x)=3#?

Question

Brooklyn Anderson · Accepted Answer

Using #sin^2(x)+cos^2(x) = 1#, we can express the equation as:

#3 = 2cos^2(x)+4sin^2(x) = 2(1-sin^2(x))+4sin^2(x)#

#=2-2sin^2(x)+4sin^2(x)#

#=2+2sin^2(x)#

Subtract 2 from both ends to get:

#2sin^2(x)=1#

Divide both sides by 2 to get:

#sin^2(x)=1/2#

Again, since #sin^2(x)+cos^2(x) = 1#, we also have:

#cos^2(x)=1-sin^2(x)=1-1/2=1/2#

Hence #sin(x)=+-sqrt(2)/2# and #cos(x)=+-sqrt(2)/2#

This is true for #x = pi/4+(n pi)/2# for all integer values of #n#.

Carter Anderson · Answer

undefined To solve the equation $2\cos^2(x) + 4\sin^2(x) = 3$, follow these steps:

1. Use the Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$.
2. Substitute $\sin^2(x) = 1 - \cos^2(x)$ into the equation.
3. Simplify the equation using the substitution and distribute the constants.
4. Combine like terms and move all terms to one side to set the equation equal to zero.
5. Factor the quadratic equation.
6. Solve for $\cos(x)$.
7. Find the solutions for $x$ by taking the inverse cosine of the solutions for $\cos(x)$.
8. Verify solutions by substituting them back into the original equation to ensure they satisfy the equation.

Following these steps, you can find the solutions for $x$ that satisfy the given equation.