# What is tanx if 2cosx sinx 1

## Equations with sine, cosine and tangent

### What is a trigonometric equation?

A **trigonometric equation** is an equation in which at least one **trigonometric function****Sine, cosine or tangent** occurs.

To solve such equations, you need a calculator. Make sure this is on *DEG* For *degree*, so **Angle measure**is set.

### Solving trigonometric equations

### $ \ sin (x) = c $

A **trigonometric equation** is given for example by $ \ sin (x) = 0.5 $. So all values for $ x $ are searched for, for which $ f (x) = \ sin (x) = 0.5 $. Look at the graph of the function $ f (x) = \ sin (x) $.

- To get a solution to the above equation, use the inverse function of $ \ sin (x) $ on the calculator, den
**Arcsine**$ \ sin ^ {- 1} $ or $ \ arcsin $. - A solution to the equation is then $ x_1 = sin ^ {- 1} (0.5) = 30 ^ \ circ $.
- The calculator always outputs values between $ -90 ^ \ circ $ and $ 90 ^ \ circ $ for equations of the form $ \ sin (x) = c $, with $ c \ in [-1; 1] $.
- As you can see from the function graph, there is another solution. You get this by subtracting the solution given by the calculator from $ 180 ^ \ circ $, i.e. $ 30 ^ \ circ $: $ x_2 = 180 ^ \ circ-30 ^ \ circ = 150 ^ \ circ $.
- The thus obtained
**Solution pair**$ x_1 = 30 ^ \ circ $ and $ x_2 = 150 ^ \ circ $ are called**Basic solution**designated. - Due to the $ 360 ^ \ circ $ -
**periodicity**the**Sine function**then all solutions of the equation are given by:

$ \ quad ~~~ x_1 ^ {(k)} = 30 ^ \ circ + k \ cdot 360 ^ \ circ $, $ k \ in \ mathbb {Z} $ and

$ \ quad ~~~ x_2 ^ {(k)} = 150 ^ \ circ + k \ cdot 360 ^ \ circ $, $ k \ in \ mathbb {Z} $.

Similarly, if there is a negative number on one side of the equation, you get all the solutions: $ \ sin (x) = - 0.5 $.

- Then $ x_1 = \ sin ^ {- 1} (- 0.5) = - 30 ^ \ circ $.
- The other
**Basic solution**then $ x_2 = -180 ^ \ circ + 30 ^ \ circ = -150 ^ \ circ $. - Here, too, you get the totality of the solution with the help of the periodicity.

$ \ quad ~~~ x_1 ^ {(k)} = -30 ^ \ circ-k \ cdot 360 ^ \ circ $, $ k \ in \ mathbb {Z} $ and

$ \ quad ~~~ x_2 ^ {(k)} = -150 ^ \ circ-k \ cdot 360 ^ \ circ $, $ k \ in \ mathbb {Z} $.

### $ \ cos (x) = c $

- The calculator always outputs values between $ 0 ^ \ circ $ and $ 180 ^ \ circ $ for equations of the form $ \ cos (x) = c $, with $ c \ in [-1; 1] $.
- The other
**Basic solution**you get by swapping the sign. - Here, too, you can use the
**periodicity**the cosine function.

**example**: $ \ cos (x) = \ frac1 {\ sqrt2} $

Then

$ x_1 = \ cos ^ {- 1} \ left (\ frac1 {\ sqrt2} \ right) = 45 ^ \ circ $.

Now $ x_2 = -45 ^ \ circ $ and

$ \ quad ~~~ x_1 ^ {(k)} = 45 ^ \ circ + k \ cdot 360 ^ \ circ $, $ k \ in \ mathbb {Z} $ and

$ \ quad ~~~ x_2 ^ {(k)} = - 45 ^ \ circ + k \ cdot 360 ^ \ circ $, $ k \ in \ mathbb {Z} $.

### $ \ tan (x) = c $

- The
**Tangent function**is $ 180 ^ \ circ $ -**periodically**. - The calculator outputs an angle between $ -90 ^ \ circ $ and $ 90 ^ \ circ $. (Note that the tangent is not defined for either $ 90 ^ \ circ $ or $ -90 ^ \ circ $.)

**example**: $ \ tan (x) = 1 $

- The calculator solution is $ x = \ tan ^ {- 1} (1) = 45 ^ \ circ $.
- The set of solutions is then given by

$ \ quad ~~~ x ^ {(k)} = 45 ^ \ circ + k \ cdot 180 ^ \ circ $, $ k \ in \ mathbb {Z} $.

### Trigonometric equations with two trigonometric functions and the same argument

How can you **trigonometric equation** solve in which two different trigonometric functions occur with the same argument?

$ (\ cos (x)) ^ 3-2 \ cos (x) \ cdot \ sin ^ 2 (x) = 0 $

- First you factor out $ \ cos (x) $.

$ \ quad ~~~ \ cos (x) \ left (\ cos ^ 2 (x) -2 \ sin ^ 2 (x) \ right) = 0 $

- A product becomes $ 0 $ when one of the factors becomes $ 0 $. So either $ \ cos (x) = 0 $ or $ \ cos ^ 2 (x) -2 \ sin ^ 2 (x) = 0 $.
- The zeros of $ \ cos (x) $ are $ x = (2k + 1) \ cdot 90 ^ \ circ $, $ k \ in \ mathbb {Z} $, i.e. the odd multiples of $ 90 ^ \ circ $.
- Now the second factor remains. Because of $ \ sin ^ 2 (x) + \ cos ^ 2 (x) = 1 $, this is the
**trigonometric pythagoras**, $ \ cos ^ 2 (x) = 1- \ sin ^ 2 (x) $ and thus

$ \ quad ~~~ 1- \ sin ^ 2 (x) -2 \ sin ^ 2 (x) = 1-3 \ sin ^ 2 (x) = 0 $.

- You can transform this equation as follows.

$ \ quad ~~~ \ begin {array} {rclll} 1-3 \ sin ^ 2 (x) & = & 0 & | & + 3 \ sin ^ 2 (x) \ 1 & = & 3 \ sin ^ 2 (x) & | &: 3 \ \ frac13 & = & \ sin ^ 2 (x) & | & \ sqrt {~~~} \ \ pm \ frac1 {\ sqrt3} & = & \ sin (x) & | & \ sin ^ {- 1} (~~~) \ \ pm35,3 ^ \ circ & \ approx & x \ end {array} $

- For each of the two solutions, as above, you can first choose the missing one
**Basic solution**determine and then the**Total solution**.

### Trigonometric equations with two trigonometric functions and different arguments

Such **equation** is given for example by

$ \ cos (x) - \ sin \ left (\ frac x2 \ right) = 0 $.

Not just two different ones dive here **Trigonometric functions** on, but also various arguments.

$ \ quad ~~~ \ cos (x) = \ cos \ left (2 \ cdot \ frac x2 \ right) $

$ \ quad ~~~ $ using a **Addition theorem** rewritten:

$ \ quad ~~~ \ cos \ left (2 \ cdot \ frac x2 \ right) = 1-2 \ sin ^ 2 \ left (\ frac x2 \ right) $.

- With that, the above equation can be written as follows:

$ \ quad ~~~ 1-2 \ sin ^ 2 \ left (\ frac x2 \ right) - \ sin \ left (\ frac x2 \ right) = 0 $

- This is a quadratic function in $ \ sin (x) $. If you

$ \ quad ~~~ z = \ sin \ left (\ frac x2 \ right) $

Substituting $ \ quad ~~~ $, you get the quadratic equation $ 1-2z ^ 2-z = 0 $.

- You can do this with the
**p-q formula**to solve. For this you put the equation around $ -2z ^ 2-z + 1 = 0 $ and divide by $ -2 $.

$ \ quad ~~~ \ begin {array} {rclll} -2z ^ 2-z + 1 & = & 0 & | &: (- 2) \ z ^ 2 + \ frac12z- \ frac12 & = & 0 \ z_ {1, 2} & = & - \ frac14 \ pm \ sqrt {\ frac1 {16} + \ frac12} \ z_1 & = & - \ frac14 + \ frac34 = \ frac12 \ z_2 & = & - \ frac14- \ frac34 = -1 \ end {array} $

- Finally, you resubstitute. So you need to solve the following equations:

$ \ quad ~~~~ \ sin \ left (\ frac x2 \ right) = \ frac12 $ and

$ \ quad ~~~~ \ sin \ left (\ frac x2 \ right) = - 1 $.

- Proceed as in the above examples for $ \ sin (x) = c $.

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