2cos^2(x) = 13sinx - 5
How do i solve for x?
I think i have to use the sin^2x + cos^2x = 1, but I'm not sure how to use that. I thought about maybe doing 2(1-sin^2x) = 13sinx - 5 but then i get stuck. Any suggestions?
2 (1-sin^2 x) = 13 sin 5 - 5
2 - 2 sin^2 x = 13 sin x - 5
2 sin^2 x + 13 sin x - 7 = 0
let z = sin x
2 z^2 + 13 z - 7 = 0
(2z -1)(z+7) = 0
sin x = 1/2 or sin x = -7
-7 is not allowed, beyond range of sin function
so x = 30 degrees
Oh that makes sense. That 2sin^2x + 13sinx - 7 can be factored. Thanks!
To solve the equation 2cos^2(x) = 13sin(x) - 5, you can indeed manipulate the equation using the trigonometric identity sin^2(x) + cos^2(x) = 1. Here's how you can proceed:
1. Start with the given equation: 2cos^2(x) = 13sin(x) - 5.
2. Rearrange the equation to isolate the cosine term: cos^2(x) = (13sin(x) - 5)/2.
3. Use the identity sin^2(x) + cos^2(x) = 1 to replace cos^2(x) in the equation: sin^2(x) + (13sin(x) - 5)/2 = 1.
4. Multiply through by 2 to eliminate the fraction: 2sin^2(x) + 13sin(x) - 5 = 2.
5. Rearrange the equation and set it to zero: 2sin^2(x) + 13sin(x) - 7 = 0.
Now you have a quadratic equation in terms of sin(x). To solve for sin(x), you can use factoring, completing the square, or the quadratic formula.
It is worth noting that solving this equation for x might not yield exact solutions, as trigonometric equations often have multiple solutions or require further approximations.
To solve the equation 2cos^2(x) = 13sinx - 5 for x, you are on the right track by considering the identity sin^2x + cos^2x = 1. Here's how you can proceed:
1. Rewrite the given equation using the identity sin^2x + cos^2x = 1:
2(1 - sin^2x) = 13sinx - 5
2. Distribute the 2 on the left side:
2 - 2sin^2x = 13sinx - 5
3. Rearrange the terms to form a quadratic equation:
2sin^2x + 13sinx - 7 = 0
Now we have a quadratic equation in terms of sinx. To solve it, you can use factoring, the quadratic formula, or completing the square. Let's use factoring in this case:
4. Factor the quadratic equation:
(sinx - 1)(2sinx + 7) = 0
Now we have two factors that, when set equal to zero, will give us potential solutions for sinx:
a) sinx - 1 = 0
Solving this equation, we get sinx = 1. Taking the inverse sine (or arcsin) of 1, we find that x = π/2 plus any multiple of 2π. So x = π/2, 5π/2, 9π/2, etc.
b) 2sinx + 7 = 0
Solving this equation, we get sinx = -7/2. However, since the range of sine function is -1 to 1, there are no real solutions for this case. Therefore, we discard this factor.
So, the solutions for x are x = π/2, 5π/2, 9π/2, etc.