Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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
3  
Does the answer sinx=1/2 sound right to everyone? –  Bilbo Dec 21 '12 at 0:16
    
ye check my solution –  MSKfdaswplwq Dec 21 '12 at 0:23

4 Answers 4

Hint: You can use the identity: $$\cos^2(x) + \sin^2x =1.$$

Using substitution, you can then obtain a quadratic equation by letting $y = \sin x$.

share|improve this answer

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

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:

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

share|improve this answer
    
+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..... –  MSKfdaswplwq Dec 21 '12 at 0:22

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

share|improve this answer
2  
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 $x$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.