I am having difficulties solving the following equation:
$4\cos^2x=5-4\sin x$
Hints on how to solve this equation would be helpful.
I am having difficulties solving the following equation:
$4\cos^2x=5-4\sin x$
Hints on how to solve this equation would be helpful.
Hint: You can use the identity: $\cos^2(x) + \sin^2x =1.$
Using substitution, you can then obtain a quadratic equation by letting $y = \sin x$.
$\cos^2(x) = 5-4 \sin(x)$
Move everything to the left hand side.
$\cos^2(x)-5+4 \sin(x) = 0$
Write in terms of $sin(x)$ using the identity $\cos^2(x) = 1-\sin^2(x)$:
$4 \sin(x)-4-\sin^2(x) = 0$
Factor constant terms from the left hand side and write the remainder as a square:
$-(\sin(x)-2)^2 = 0$
Multiply both sides by -1:
$(\sin(x)-2)^2 = 0$
Take the square root of both sides:
$\sin(x)-2 = 0$
Add 2 to both sides: $\sin(x) = 2$
Your edited your post:
$4 cos^2(x) = 5-4 sin(x)$
Subtract $5-4 sin(x)$ from both sides:
$4 cos^2(x)-5+4 sin(x) = 0$
Using the identity $cos^2(x) = 1-sin^2(x):$
$4 sin(x)-1-4 sin^2(x) = 0$
Factor constant terms from the left hand side and write the remainder as a square:
$-(2 sin(x)-1)^2 = 0$
Multiply both sides by -1:
$(2 sin(x)-1)^2 = 0$
Take the square root of both sides:
$2 sin(x)-1 = 0$
Add 1 to both sides:
$2 sin(x) = 1$
Divide both sides by 2:
$sin(x) = 1/2$
Hint: Substitute in $\cos^2x=1-\sin^2x$, and solve the quadratic for $\sin x$.
I am having difficulties solving the following equation:${4\cos^2x=5-4\sin x}$
First, substitute $4\cos^2(x)$ with $4\left(1 - \sin^2(x)\right) = 4 - 4\sin^2(x).$ We are left with $4 - 4\sin^2(x) = 5 - 4 \sin (x) .$ This can be rewritten as $-4\sin^2(x) + 4\sin(x) - 1 = 0.$
Observe that the equation can further be rewritten in the form $-4t^2 + 4t - 1 = 0$ where $t = \sin(x)$. Solve the quadratic equation for $t$ and then use $\sin(x) = t$ to solve for $x$.