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Question: How do you solve \(3\sin x + 5\cos x = 4\) ?...

How do you solve 3sinx+5cosx=43\sin x + 5\cos x = 4 ?

Explanation

Solution

First substitute sinx=u\sin x = u and cosx=v\cos x = v. Then isolate and make the equation in terms of variable uu. Use the trigonometric identity sin2x+cos2x=1{\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: sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1, b±b24ac2a\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 3sinx+5cos=43\sin x + 5\cos = 4
Now, in order to solve this, let us consider sinx=u\sin x = u and cosx=v\cos x = v
According to the question:
3u+5v=4\Rightarrow 3u + 5v = 4
Adding 5v - 5v to both sides of the equation, we get:
3u=45v\Rightarrow 3u = 4 - 5v
Dividing both sides of the equation with 33, we get:
u=45v3\Rightarrow u = \dfrac{{4 - 5v}}{3}
Now according to the common trigonometric identity, we know that:
sin2x+cos2x=1\Rightarrow {\sin ^2}x + {\cos ^2}x = 1
Therefore,u2+v2=1{u^2} + {v^2} = 1
Let us substitute the value of uu that we have calculated prior, in the above equation:
(45v3)2+v2=1\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:
(45v)29+v2=1\Rightarrow \dfrac{{{{\left( {4 - 5v} \right)}^2}}}{9} + {v^2} = 1
As we know that, (ab)2=a22ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}
Therefore, 1640v+25v29+v2=1\dfrac{{16 - 40v + 25{v^2}}}{9} + {v^2} = 1
Taking LCM on the left hand side, we get:
1640v+25v2+9v29=1\Rightarrow \dfrac{{16 - 40v + 25{v^2} + 9{v^2}}}{9} = 1
Multiplying both sides of the equation with 99 :
1640v+34v2=9\Rightarrow 16 - 40v + 34{v^2} = 9
On adding 9 - 9 to both sides, we get:
1640v+34v29=0\Rightarrow 16 - 40v + 34{v^2} - 9 = 0
40v+34v2+7=0\Rightarrow - 40v + 34{v^2} + 7 = 0
On rearranging the terms, we get:
34v240v+7=0\Rightarrow 34{v^2} - 40v + 7 = 0
Now we have a quadratic equation. We can solve this equation using the quadratic formula which is b±b24ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
If we compare our quadratic function with the standard quadratic function ax2+bx+c=0a{x^2} + bx + c = 0 , we find that
a=34a = 34
b=40b = - 40
C=77
Now let us find the roots of our quadratic equation:
v=b±b24ac2av = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
On substituting the values, we get:
v=(40)±(40)24×34×72×34v = \dfrac{{ - \left( { - 40} \right) \pm \sqrt {{{\left( { - 40} \right)}^2} - 4 \times 34 \times 7} }}{{2 \times 34}}
v=40±160095268\Rightarrow v = \dfrac{{40 \pm \sqrt {1600 - 952} }}{{68}}
On simplifying it further, we get:
v=40±64868v = \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=40+25.4568v = \dfrac{{40 + 25.45}}{{68}} or v=4025.4568v = \dfrac{{40 - 25.45}}{{68}}
v=65.4568\Rightarrow v = \dfrac{{65.45}}{{68}} or v=14.5568v = \dfrac{{14.55}}{{68}}
v=0.9625\Rightarrow v = 0.9625 or v=0.2139v = 0.2139
Now, as we calculated earlier:
u=45v3u = \dfrac{{4 - 5v}}{3}
Thus there can also be two values of uu
When v=0.9625v = 0.9625
u=45v3=0.81253=0.2708u = \dfrac{{4 - 5v}}{3} = \dfrac{{ - 0.8125}}{3} = - 0.2708
When v=0.2139v = 0.2139
u=45v3=2.93053=0.9768u = \dfrac{{4 - 5v}}{3} = \dfrac{{2.9305}}{3} = 0.9768

**Therefore, sinx=0.2708\sin x = - 0.2708 or 0.97680.9768
And cosx=0.9625\cos x = 0.9625 or 0.21390.2139 **

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:
sinθ=perpendicularhypotenuse{{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.
cosθ=basehypotenuse{{cos\theta = }}\dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}} , where base is the side containing the perpendicular and hypotenuse
tanθ=perpendicularbase{{tan\theta = }}\dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}