Solving Trigonometric Equations: A Step-by-Step Guide

Hey guys! Let's dive into the world of trigonometry and tackle some tricky equations. Solving trigonometric equations can seem daunting at first, but with a systematic approach and a little practice, you'll be acing these problems in no time. We'll go through each equation step-by-step, explaining the reasoning behind each move. So, grab your pencils and let's get started! This guide will provide you with the necessary tools and knowledge to confidently solve a variety of trigonometric equations. We'll be looking at different types of equations, from quadratic-like forms to those involving multiple trigonometric functions. Ready to become a trigonometry whiz?

Equation 1: 4sin²x + sinx - 5 = 0

Alright, let's start with our first equation: 4sin²x + sinx - 5 = 0. This looks a lot like a quadratic equation, doesn't it? That's because we can treat sinx as our variable. The key here is recognizing the quadratic form and making a substitution to simplify things. Let's make it easier to understand. This equation is actually a quadratic equation in terms of sin(x). We can rewrite this equation as 4(sin x)^2 + sin x - 5 = 0. See how it looks like a quadratic equation now? Let's use substitution. This is one of the most useful tricks in algebra. Let's substitute y = sin x. Therefore, our equation now becomes:

• 4y² + y - 5 = 0

Now, we can solve this quadratic equation using the quadratic formula, factoring, or any method you prefer. Let's try factoring it. So, let's factorize 4y² + y - 5 = 0. We're looking for two numbers that multiply to give us -20 (4 * -5) and add up to 1. Those numbers are 5 and -4. So we split the middle term and factor by grouping: 4y² - 4y + 5y - 5 = 0 => 4y(y-1) + 5(y-1) = 0 => (4y + 5)(y - 1) = 0. This gives us two possible solutions for y:

• 4y + 5 = 0 => y = -5/4
• y - 1 = 0 => y = 1

Remember, we substituted y = sin x. So now we need to substitute back to find the values of x. Let's work through this. For y = -5/4, which is equivalent to sin x = -5/4. However, the range of the sine function is [-1, 1]. This means that the value of sin x can never be less than -1 or greater than 1. Thus, sin x = -5/4 has no solution. Now, let's look at y = 1, which means sin x = 1. The sine function equals 1 at x = π/2 + 2πn, where n is an integer. Thus, the solution to the equation is x = π/2 + 2πn. Congratulations! You've successfully solved your first trigonometric equation.

Equation 2: 6sin²x + 7cosx - 1 = 0

On to the next challenge! This time we have 6sin²x + 7cosx - 1 = 0. This equation involves both sine and cosine functions. Our strategy here is to rewrite the equation so that it only involves one trigonometric function. In this case, we can use the Pythagorean identity: sin²x + cos²x = 1. Therefore sin²x = 1 - cos²x. Let's make the substitution. Substituting sin²x = 1 - cos²x into our equation, we get:

• 6(1 - cos²x) + 7cosx - 1 = 0

Now, let's simplify and rearrange the equation. Expanding the equation, we get: 6 - 6cos²x + 7cosx - 1 = 0 => -6cos²x + 7cosx + 5 = 0. Let's multiply both sides by -1 to make the leading coefficient positive. The equation now becomes: 6cos²x - 7cosx - 5 = 0. Now, let's substitute y = cos x to make it more simple. Thus our equation will now be: 6y² - 7y - 5 = 0. This is a quadratic equation. Let's factorize. We are looking for two numbers that multiply to give us -30 (6 * -5) and add to -7. Those numbers are -10 and 3. We split the middle term, factor by grouping: 6y² - 10y + 3y - 5 = 0 => 2y(3y - 5) + 1(3y - 5) = 0 => (2y + 1)(3y - 5) = 0. Thus, we get two possible solutions for y:

• 2y + 1 = 0 => y = -1/2
• 3y - 5 = 0 => y = 5/3

Now we substitute back cos x for y. We get cos x = -1/2 and cos x = 5/3. As we know, the range of the cosine function is [-1, 1]. So, cos x = 5/3 has no solution. For cos x = -1/2, the solutions are x = 2π/3 + 2πn and x = 4π/3 + 2πn, where n is an integer. Awesome job! You're really getting the hang of this.

Equation 3: 4sin²x + sinxcosx + 6cos²x = 0

Now, let's analyze 4sin²x + sinxcosx + 6cos²x = 0. This one looks a bit different, but don't worry, we can handle it. The main idea here is to divide the entire equation by cos²x (assuming cos x ≠ 0). This will transform the equation into terms of tan x. So, let's divide each term by cos²x:

• (4sin²x / cos²x) + (sinxcosx / cos²x) + (6cos²x / cos²x) = 0

This simplifies to:

• 4tan²x + tanx + 6 = 0

Let's substitute y = tan x. Then we will have: 4y² + y + 6 = 0. Let's find the discriminant to check if there are any real solutions. The discriminant is b² - 4ac, which is equal to 1² - 4 * 4 * 6 = 1 - 96 = -95. Since the discriminant is negative, this quadratic equation has no real solutions. Thus, the original equation has no real solutions. This is an important reminder: always check the validity of your solutions, especially in trigonometric equations, as extraneous solutions can arise.

Equation 4: 5tanx - 6cotx + 13 = 0

Here we go with 5tanx - 6cotx + 13 = 0. This equation involves tangent and cotangent functions. Remember that cotx = 1/tanx. Let's rewrite the equation in terms of tanx. Let's substitute y = tanx. The equation then becomes:

• 5y - 6/y + 13 = 0

Now, let's multiply the whole equation by y to eliminate the fraction (note that y ≠ 0): 5y² - 6 + 13y = 0. Let's rearrange: 5y² + 13y - 6 = 0. Factorize this equation. We're looking for two numbers that multiply to give us -30 (5 * -6) and add up to 13. These numbers are 15 and -2. We split the middle term and factor by grouping: 5y² + 15y - 2y - 6 = 0 => 5y(y + 3) - 2(y + 3) = 0 => (5y - 2)(y + 3) = 0. Now we get the solutions for y:

• 5y - 2 = 0 => y = 2/5
• y + 3 = 0 => y = -3

Now let's substitute back tanx for y. We get tanx = 2/5 and tanx = -3. The solutions for tanx = 2/5 are x = arctan(2/5) + πn, where n is an integer. The solutions for tanx = -3 are x = arctan(-3) + πn, where n is an integer. Keep up the good work!

Equation 5: 3 - 4sin²x = sin2x

Next up: 3 - 4sin²x = sin2x. This equation involves sin²x and sin2x. Let's use the double-angle identity: sin2x = 2sinxcosx. Let's rearrange the equation and substitute the double-angle identity. Let's replace sin2x with 2sinxcosx. Our equation now becomes: 3 - 4sin²x = 2sinxcosx. We can rewrite the equation as 4sin²x + 2sinxcosx - 3 = 0. This form doesn't immediately suggest a clear path to a solution, but we can look for other trigonometric identities to help us. Let's apply sin²x + cos²x = 1, so we have 3 = 3(sin²x + cos²x). Let's substitute and rearrange the equation: 4sin²x + 2sinxcosx - 3(sin²x + cos²x) = 0. Then simplify: 4sin²x + 2sinxcosx - 3sin²x - 3cos²x = 0 => sin²x + 2sinxcosx - 3cos²x = 0. Now, divide everything by cos²x (assuming cosx ≠ 0): (sin²x / cos²x) + (2sinxcosx / cos²x) - (3cos²x / cos²x) = 0. Which simplifies to tan²x + 2tanx - 3 = 0. Let's substitute y = tan x. Our equation will become: y² + 2y - 3 = 0. Factorize this equation: (y + 3)(y - 1) = 0. We get two possible solutions: y = -3 and y = 1. Substitute tanx for y. Then, we have tanx = -3 and tanx = 1. The solution for tanx = -3 are x = arctan(-3) + πn, where n is an integer. The solution for tanx = 1 are x = π/4 + πn, where n is an integer. You are doing great!

Equation 6: 10sin2x + 3cos2x = -3 - 14sin²x

Finally, let's crack 10sin2x + 3cos2x = -3 - 14sin²x. This one is a bit more involved. We'll utilize multiple trigonometric identities to solve it. Let's start with the double-angle identities: sin2x = 2sinxcosx and cos2x = 1 - 2sin²x. Let's substitute: 10(2sinxcosx) + 3(1 - 2sin²x) = -3 - 14sin²x. Expand the equation: 20sinxcosx + 3 - 6sin²x = -3 - 14sin²x. Let's rearrange: 20sinxcosx + 8sin²x + 6 = 0. We can simplify it by dividing by 2: 10sinxcosx + 4sin²x + 3 = 0. Let's substitute 3 = 3(sin²x + cos²x), and then our equation will look like this: 10sinxcosx + 4sin²x + 3sin²x + 3cos²x = 0 => 10sinxcosx + 7sin²x + 3cos²x = 0. Divide by cos²x (assuming cosx ≠ 0): (10sinxcosx / cos²x) + (7sin²x / cos²x) + (3cos²x / cos²x) = 0. That gives us 10tanx + 7tan²x + 3 = 0. Reorder it as 7tan²x + 10tanx + 3 = 0. Let's substitute y = tanx. We get 7y² + 10y + 3 = 0. Factorize: (7y + 3)(y + 1) = 0. Then, we get y = -3/7 and y = -1. For y = -3/7, so tanx = -3/7. This gives us x = arctan(-3/7) + πn. For y = -1, so tanx = -1, which gives us x = 3π/4 + πn. Congratulations! You've successfully solved all the equations!

And that's a wrap! You've just solved six different trigonometric equations. Remember, practice is key. The more you solve these problems, the more familiar you'll become with the different techniques and identities. Keep practicing and exploring, and you'll become a trigonometry master. Good luck, and keep learning!"