# Help Solving Trigonometry Equation

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.

-
Try adding dollar signs around your equation, and back slashs before the cos and sin, to make the mathematics appear more visually appealing. For the above, I changed the equation to $$4\cos^2x=5-4\sin x.$$ This yields $$4\cos^2x=5-4\sin x.$$ – Eric Naslund Dec 21 '12 at 0:04
Hint: Do you know a trig identity you could substitute in for $cos^{2}x$ that use $sin^{2}x$? Then simplify and solve the remaining equation. – Amzoti Dec 21 '12 at 0:09
Does the answer sinx=1/2 sound right to everyone? – Bilbo Dec 21 '12 at 0:16
ye check my solution – Applied mathematician Dec 21 '12 at 0:23

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: $\sin(x) = 2$

$$4 cos^2(x) = 5-4 sin(x)$$

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:

$$2 sin(x)-1 = 0$$

$$2 sin(x) = 1$$

Divide both sides by 2:

$$sin(x) = 1/2$$

-
+1, Interesting, I like that you provided a complete solution to a different problem than what the OP was asking. This gives a strong hint. – Eric Naslund Dec 21 '12 at 0:20
It was not my intention :) Bilbo made a mistake. Now I gave him the correct solution..... – Applied mathematician Dec 21 '12 at 0:22

Hint: Substitute in $\cos^2x=1-\sin^2x$, and solve the quadratic for $\sin x$.

-
I always forget to use factoring as an option to solve when solving trigonometry equations, thank you. – Bilbo Dec 21 '12 at 0:11

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 $x$.

-