Please take a look at DEVON's trig equation at 9:14
I must be missing something obvious.
If the second had been 3sin 2x it would be a straight forward question.
The way it stands I was able to "reduce" it to the equation
8sin^4 x - 8sin^2 x + 3sin x + 2 = 0
I tested this equation by subst. random values of angles in the original and this one, and got the same results, so it must be right.
??????
(must have been infected with a case of "oldtimers' disease)
2cos^2x+3sinx=0
2-2sin^2 x + 3 sinx=0
sin^2 x - 3/2 sin x - 1=0
quadratic equation..
sinx= (3/2 +-sqrt(9/4 +4) /2
sinx= 3/4 +- 5/4
sin x= -1/2 works.
check me.
Of course, of course, I knew it had to be that easy
I read the exponent as a multiplier of x
i.e. 2 cos^2 (2x) + 3 sin x .....
To solve the trigonometric equation 2cos^2x + 3sinx = 0, we can follow these steps:
1. Use the identity cos^2x = 1 - sin^2x to rewrite the equation: 2(1 - sin^2x) + 3sinx = 0.
2. Simplify the equation: 2 - 2sin^2x + 3sinx = 0.
3. Rearrange the terms: -2sin^2x + 3sinx + 2 = 0.
4. Now, you have correctly rewritten the equation as 8sin^4x - 8sin^2x + 3sinx + 2 = 0. This is a quadratic equation in terms of sinx.
5. Solve the quadratic equation using factoring, quadratic formula, or completing the square methods. In this case, factoring might be the easiest approach.
You have correctly further simplified the equation to sin^2x - (3/2)sinx - 1 = 0.
To solve this quadratic equation, you can either factor it or use the quadratic formula. Let's use the quadratic formula:
1. Identify the coefficients of the quadratic equation: a = 1, b = -(3/2), and c = -1.
2. Apply the quadratic formula: sinx = (-b ± √(b^2 - 4ac)) / (2a).
3. Substitute the values into the quadratic formula: sinx = (-(3/2) ± √((3/2)^2 - 4(1)(-1))) / (2(1)).
4. Simplify the expression: sinx = (-(3/2) ± √(9/4 + 4)) / 2.
This becomes sinx = (-(3/2) ± √(25/4)) / 2.
5. Continue simplifying: sinx = (-(3/2) ± (5/2)) / 2.
This gives two possible solutions: sinx = (1/2) or sinx = -2.
6. Check the solution sinx = -1/2:
Substitute sinx = -1/2 into the original equation to verify if it satisfies the equation. If it does, then it is a valid solution.
Plugging in sinx = -1/2 into the equation 2cos^2x + 3sinx = 0, we get:
2cos^2x + 3(-1/2) = 0
Simplifying further, we have 2cos^2x - (3/2) = 0.
Rearranging the terms, we get 2cos^2x = (3/2).
Dividing by 2, we have cos^2x = (3/4).
Taking the square root, we get cosx = ±√(3/4), which simplifies to ±√3/2.
So, when sinx = -1/2, we have two possible solutions: cosx = √3/2 and cosx = -√3/2.
Thus, the solutions to the equation 2cos^2x + 3sinx = 0 are:
1. sinx = 1/2, cosx = √3/2
2. sinx = -1/2, cosx = √3/2
3. sinx = -1/2, cosx = -√3/2
You have correctly identified sinx = -1/2 as a valid solution, and you can check the corresponding values of cosx to verify.