# Solving Trigonometric Equations

To become an effective trigonometric equations solver, we need to know the basic trigonometric identities. I created a guideline below to follow so that it would be easy for you where to start solving trigonometric equations.

Guidelines:

1. Familiarize and memorize these identities. That is your foundation.
2. Use these identity to solve the equations given by expressing the it to one unknown
3. If the degree of equations derived is more than 1, use quadratic formula or factoring to simplify it.

Worked Problem 1:

Find the value of x in $2sinx-\sqrt{3}=0$ if $0\le x\le \pi$.

Solution:

$2sinx-\sqrt{3}=0$

$2sinx=\sqrt{3}$

$sinx=\displaystyle\frac{\sqrt{3}}{2}$

$\displaystyle\frac{\sqrt{3}}{2}$ and 0.5 are special numbers in trigonometry. Because they are coordinates of special angles like 60° and 30°.

We know that sin30°=0.5. This means that

$sin60=\displaystyle\frac{\sqrt{3}}{2}$

But it’s not just 60. 120° is also a solution which is also 60° from x-axis.

Since the interval given is in radians, we need to express our answer also in radians. Thus the values of x are $\displaystyle\frac{\pi}{3} and \displaystyle\frac{2\pi}{3}$

Sample Problem 2:

Find x in $4cos^2x-1=0$ if $0\le c\le 360$°.

Solution:

$4cos^2x-1=0$

$4cos^2x=1$

$cos^2x=\displaystyle\frac{1}{4}$

$\sqrt{cos^2x}=\sqrt{\displaystyle\frac{1}{4}}$

$cosx=\pm\displaystyle\frac{1}{2}$

½ is the coordinate of cosine of all angles 60° from x-axis. Those are 60°,120°,240°, and 300°

Sample Problem 3:

Solve for x in $5-4cos^2x-4sinx=0$  in the interval (0,2π).

Solution:

Now, in this problem we have two unknowns. This is where our skill in identity can be harnessed.

From Pythagorean Idenitity,

$sin^2x+cos^2x=1$

$cos^2x=1-sin^2x$

Substitute this to the given equation so that we will have only one unknown.

$5-4cos^2x-4sinx=0$

$5-4(1-sin^2x)-4sinx=0$

$5-4-4sin^2x-4sinx=0$

$1-4sinx-4sin^2x=0$

By factoring,

$(1-2sinx)^2=0$

$1-2sinx=0$

$1=2sinx$

$sinx=\displaystyle\frac{1}{2}$

$x=sin^{-1}\displaystyle\frac{1}{2}$

$x=\displaystyle\frac{pi}{6},\displaystyle\frac{5pi}{6}$

Sample Problem 4:

Find the value of x if  $sin2xcosx-cosxsin2x=\displaystyle\frac{1}{2}$   in the interval (0,90°)

Solution:

Using the sum and difference identity sin(x-y)=sinxcosy-cosxsiny

$sin2xcosx-cosxsin2x=sin(2x-x)=\displaystyle\frac{1}{2}$

$sin(2x-x)=\displaystyle\frac{1}{2}$

$sin(x)=\displaystyle\frac{1}{2}$

$x=sin^{-1}\displaystyle\frac{1}{2}$

$x=30$°

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

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