Question

How do you solve $3\sin x + 5\cos x = 4$ ?

Answer 1

Hint: First substitute $\sin x = u$ and $\cos x = v$. Then isolate and make the equation in terms of variable $u$. Use the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$ and substitute the respective values. Solve it further and simplify to get a quadratic function. Solve the quadratic function using the quadratic formula to get the respective roots of the question.

Formula used: ${\sin ^2}x + {\cos ^2}x = 1$, $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
In this question, we are asked to solve the trigonometric expression given as $3\sin x + 5\cos = 4$
Now, in order to solve this, let us consider $\sin x = u$ and $\cos x = v$
According to the question:
$ \Rightarrow 3u + 5v = 4$
Adding $ - 5v$ to both sides of the equation, we get:
\[ \Rightarrow 3u = 4 - 5v\]
Dividing both sides of the equation with $3$, we get:
$ \Rightarrow u = \dfrac{{4 - 5v}}{3}$
Now according to the common trigonometric identity, we know that:
$ \Rightarrow {\sin ^2}x + {\cos ^2}x = 1$
Therefore,${u^2} + {v^2} = 1$
Let us substitute the value of $u$ that we have calculated prior, in the above equation:
$ \Rightarrow {\left( {\dfrac{{4 - 5v}}{3}} \right)^2} + {v^2} = 1$ ,
Now our equation has become homogenous which means that it contains only one variable.
Solving it further, we find:
$ \Rightarrow \dfrac{{{{\left( {4 - 5v} \right)}^2}}}{9} + {v^2} = 1$
As we know that, ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
Therefore, $\dfrac{{16 - 40v + 25{v^2}}}{9} + {v^2} = 1$
Taking LCM on the left hand side, we get:
$ \Rightarrow \dfrac{{16 - 40v + 25{v^2} + 9{v^2}}}{9} = 1$
Multiplying both sides of the equation with $9$ :
$ \Rightarrow 16 - 40v + 34{v^2} = 9$
On adding $ - 9$ to both sides, we get:
$ \Rightarrow 16 - 40v + 34{v^2} - 9 = 0$
$ \Rightarrow - 40v + 34{v^2} + 7 = 0$
On rearranging the terms, we get:
$ \Rightarrow 34{v^2} - 40v + 7 = 0$
Now we have a quadratic equation. We can solve this equation using the quadratic formula which is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
If we compare our quadratic function with the standard quadratic function $a{x^2} + bx + c = 0$ , we find that
$a = 34$
$b = - 40$
C=$7$
Now let us find the roots of our quadratic equation:
$v = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
On substituting the values, we get:
$v = \dfrac{{ - \left( { - 40} \right) \pm \sqrt {{{\left( { - 40} \right)}^2} - 4 \times 34 \times 7} }}{{2 \times 34}}$
$ \Rightarrow v = \dfrac{{40 \pm \sqrt {1600 - 952} }}{{68}}$
On simplifying it further, we get:
$v = \dfrac{{40 \pm \sqrt {648} }}{{68}}$
Now, there can be two roots. We can take the positive value of the square root or negative value. Thus,
$v = \dfrac{{40 + 25.45}}{{68}}$ or $v = \dfrac{{40 - 25.45}}{{68}}$
$ \Rightarrow v = \dfrac{{65.45}}{{68}}$ or $v = \dfrac{{14.55}}{{68}}$
$ \Rightarrow v = 0.9625$ or $v = 0.2139$
Now, as we calculated earlier:
$u = \dfrac{{4 - 5v}}{3}$
Thus there can also be two values of $u$
When $v = 0.9625$
$u = \dfrac{{4 - 5v}}{3} = \dfrac{{ - 0.8125}}{3} = - 0.2708$
When $v = 0.2139$
$u = \dfrac{{4 - 5v}}{3} = \dfrac{{2.9305}}{3} = 0.9768$

Therefore, $\sin x = - 0.2708$ or $0.9768$
And $\cos x = 0.9625$ or $0.2139$

Note: Trigonometry is a branch of mathematics which deals with triangles. There are many trigonometric formulas that establish a relation between the lengths and angles of respective triangles. In trigonometry, we use a right-angled triangle to find ratios of its different sides and angles such as sine, cosine, tan, and their respective inverse like cosec, sec, and cot. Some common formulas of trigonometric identities are:
${{sin\theta = }}\dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}$ , where perpendicular is the side containing the right angle in a right angled triangle and hypotenuse is the side opposite to the perpendicular.
${{cos\theta = }}\dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}$ , where base is the side containing the perpendicular and hypotenuse
${{tan\theta = }}\dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$