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 functionSine, cosine or tangent occurs.

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