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## Solving second degree trigonometric equations

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- 6 sin 2 x + 3 sin x = 0 6 \sin^2x + 3 \sin x = 0 6 sin 2 x + 3 sin x = 0 for 0 ≤ x ≤ 2 π 0 \leq x \leq 2\pi 0 ≤ x ≤ 2 π
- 2 cos 2 x + 3 cos x + 1 = 0 2\cos^2 x + 3 \cos x + 1 = 0 2 cos 2 x + 3 cos x + 1 = 0 general solution in radians
- 3 sin 2 x − 8 sin x = − 4 3 \sin^2 x - 8 \sin x = -4 3 sin 2 x − 8 sin x = − 4 general solution in degrees
- 4 cos 2 x − 3 = 0 4 \cos^2 x - 3 = 0 4 cos 2 x − 3 = 0 general solution in radians
- sec 2 x − 3 sec x + 2 = 0 \sec^2x - 3 \sec x + 2= 0 sec 2 x − 3 sec x + 2 = 0 for 0 ≤ x 0 \leq x 0 ≤ x 2 π 2\pi 2 π
- Determine a cosine equation that has the following general solution: π 2 + n π , π 6 + 2 n π , 11 π 6 + 2 n π , {\pi \over 2} + n\pi , { \pi \over 6} + 2n\pi, {{11\pi} \over 6} + 2n\pi, 2 π + nπ , 6 π + 2 nπ , 6 11 π + 2 nπ , where n n n is an integer A) cos x ( 2 cos x + 2 ) = 0 \cos x (2\cos x + \sqrt2) = 0 cos x ( 2 cos x + 2 ) = 0 B) cos x ( 2 cos x + 3 ) = 0 \cos x (2 \cos x + \sqrt3) = 0 cos x ( 2 cos x + 3 ) = 0 C) cos x ( 2 cos x − 2 ) = 0 \cos x (2 \cos x - \sqrt2) = 0 cos x ( 2 cos x − 2 ) = 0 D) cos x ( 2 cos x − 3 ) = 0 \cos x (2 \cos x - \sqrt3) = 0 cos x ( 2 cos x − 3 ) = 0

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## Topic Notes

sin 2 x + sin x = 4 \sin ^2x+\sin x =4 sin 2 x + sin x = 4

cos 2 y − 4 cos y + 3 = 0 \cos ^2y-4 \cos y +3 =0 cos 2 y − 4 cos y + 3 = 0

## Solving Trigonometric Equations

Step 1: Replace sinx with another variable

Let w = sin x w = \sin x w = sin x

6 w 2 + 3 w = 0 6w^2+3w=0 6 w 2 + 3 w = 0

Step 2: Factor the newly-formed polynomial

3 w ( 2 w + 1 ) = 0 3w(2w+1)=0 3 w ( 2 w + 1 ) = 0

Hopefully you remember how to solve factored polynomials. In this case, w = 0 and w = -1/2

Step 4: Replace w with sinx and change solutions

Therefore: sin x = 0 and sin x = -1/2

For practice solving these kinds of calculations, check out the lesson on finding trig values .

Step 5: Find solutions within entire specified range

Therefore, our final answer is: x = 0, pi, 7pi/6, 11pi/6, and 2pi

2 cos 2 x + 3 c o s x + 1 = 0 2 \cos^2 x + 3\ cos x +1=0 2 cos 2 x + 3 cos x + 1 = 0

Step 1: Factor and simplify the equation

( 2 cos x + 1 ) ( cos x + 1 ) = 0 (2 \cos x+1) ( \cos x +1)=0 ( 2 cos x + 1 ) ( cos x + 1 ) = 0

Now that we've factored our equation, it's easy to solve and determine that:

Solving for x gives us the solutions: x = 2pi/3 and pi

Step 3: Find solutions within entire specified range

Therefore, our final answer is: thus x = pi + n2pi, x = 2pi/3 = n2pi, and x = 4pi/3 + n2pi

## Basic Concepts

- Reference angle
- Find the exact value of trigonometric ratios
- ASTC rule in trigonometry ( A ll S tudents T ake C alculus)
- Factoring trinomials

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## Trigonometric equations online calculator

The general form of trigonometric equation is

where - some arbitrary function, trig ( x ) - some trigonometric function.

For example, consider the solution of trigonometric equation

By using the cosine double angle formula, transform this equation to the simplier one:

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Because of the presence of these trigonometric functions, solving these equations becomes a little more difficult. Now that we have solved the equation for w, we have to remember to replace w back with sin x and adjust our solutions to accommodate this

Online calculator is able to solve any trigonometric equation with step by step solution

One way to solve a trigonometric equation is to graph the functions on each side of the equals sign,then find the point(s) of intersection. To solve, start by turning on the two equations in boxes 3 & 4 below to see the graphs