# a cos θ + b sin θ = c

Trigonometric equations of the form a cos theta plus b sin theta equals c (i.e. a cos θ + b sin θ = c) where a, b, c are constants (a, b, c ∈ R) and |c| ≤ $$\sqrt{a^{2} + b^{2}}$$.

To solve this type of questions, we first reduce them in the form cos θ = cos α or sin θ = sin α.

We use the following ways to solve the equations of the form a cos θ + b sin θ = c.

(i) First write the equation a cos θ + b sin θ = c.

(ii) Let a = r cos ∝ and b = r sin ∝ where, r > 0 and - $$\frac{π}{2}$$ ≤ ∝ ≤ $$\frac{π}{2}$$.

Now, a$$^{2}$$ + b$$^{2}$$ = r$$^{2}$$ cos$$^{2}$$ ∝ + r$$^{2}$$ sin$$^{2}$$ ∝ = r$$^{2}$$ (cos$$^{2}$$ ∝ + sin$$^{2}$$ ∝) = r$$^{2}$$

or, r =  $$\sqrt{a^{2} + b^{2}}$$

and tan ∝ = $$\frac{r sin ∝}{r cos ∝}$$ = $$\frac{b}{a}$$ i.e. ∝ = tan$$^{-1}$$ ($$\frac{b}{a}$$).

(iii) Using the substitution in step (ii), the equation reduce to r cos (θ - ∝) = c

⇒ cos  (θ - ∝) = $$\frac{c}{r}$$ = cos β

Now, putting the value of a and b in a cos θ + b sin θ = c we get,

r cos ∝ cos θ + r sin ∝ sin θ = c

⇒ r cos (θ - ∝) = c

⇒ cos (θ - ∝) = $$\frac{c}{r}$$ = cos β          (say)

(iv) Solve the equation obtained in step (iii) by using the formula of cos θ = cos ∝.

cos (θ - ∝) = cos β

Therefore, θ - ∝ = 2nπ ± β

⇒ θ = 2nπ ± β + ∝ where n ∈ Z

and cos β = $$\frac{c}{r}$$ = $$\frac{c}{\sqrt{a^{2} + b^{2}}}$$

Note: If |c| > $$\sqrt{a^{2} + b^{2}}$$, the given equation has no solution.

From the above discussion we observe that a cos θ + b sin θ = c can be solved  when |cos β| ≤ 1

⇒ |$$\frac{c}{\sqrt{a^{2} + b^{2}}}$$| ≤ 1

⇒ |c| ≤  $$\sqrt{a^{2} + b^{2}}$$

1. Solve the trigonometric equation √3 cos θ + sin θ = √2.

Solution:

√3 cos θ + sin θ = √2

This trigonometric equation is of the form a cos θ + b sin θ = c where a = √3, b = 1 and c = √2.

Let a  = r cos ∝ and b = r sin ∝ i.e., √3 = r cos ∝ and 1 = r sin ∝.

Then r = $$\sqrt{a^{2} + b^{2}}$$ = $$\sqrt{(√3)^{2} + 1^{2}}$$ = 2

and tan ∝ = $$\frac{1}{√3}$$ ∝ = $$\frac{π}{6}$$

Substituting a = √3 = r cos ∝ and b = 1 = r sin ∝ in the given equation √3 cos θ + sin θ = √2 we get,

r cos ∝ cos θ + r sin ∝ sin θ = √2

r cos (θ - ∝) = √2

⇒ 2 cos (θ - $$\frac{π}{6}$$) = √2

⇒ cos (θ - $$\frac{π}{6}$$) = $$\frac{√2}{2}$$

⇒ cos (θ - $$\frac{π}{6}$$) = $$\frac{1}{√2}$$

cos (θ - $$\frac{π}{6}$$) = cos $$\frac{π}{4}$$

(θ - $$\frac{π}{6}$$)= 2nπ ± $$\frac{π}{4}$$, where  n = 0, ± 1, ± 2,…………

θ = 2nπ ± $$\frac{π}{4}$$ + $$\frac{π}{6}$$, where  n = 0, ± 1, ± 2,…………

θ = 2nπ + $$\frac{π}{4}$$ + $$\frac{π}{6}$$ or θ = 2nπ - $$\frac{π}{4}$$ + $$\frac{π}{6}$$, where  n = 0, ± 1, ± 2,…………

θ = 2nπ + $$\frac{5π}{12}$$ or θ = 2nπ - $$\frac{π}{12}$$, where  n = 0, ± 1, ± 2,…………

2. Solve √3 cos θ + sin θ = 1  (-2π < θ < 2π)

Solution:

√3 cos θ + sin θ = 1

This trigonometric equation is of the form a cos θ + b sin θ = c where a = √3, b = 1 and c = 1.

Let a  = r cos ∝ and b = r sin ∝ i.e., √3 = r cos ∝ and 1 = r sin ∝.

Then r = $$\sqrt{a^{2} + b^{2}}$$ = $$\sqrt{(√3)^{2} + 1^{2}}$$ = 2

and tan ∝ = $$\frac{1}{√3}$$ ∝ = $$\frac{π}{6}$$

Substituting a = √3 = r cos ∝ and b = 1 = r sin ∝ in the given equation √3 cos θ + sin θ = √2 we get,

r cos ∝ cos θ + r sin ∝ sin θ = 1

r cos (θ - ∝) = 1

⇒ 2 cos (θ - $$\frac{π}{6}$$) = 1

⇒ cos (θ - $$\frac{π}{6}$$) = $$\frac{1}{2}$$

cos (θ - $$\frac{π}{6}$$) = cos $$\frac{π}{3}$$

(θ - $$\frac{π}{6}$$)= 2nπ ± $$\frac{π}{3}$$, where  n = 0, ± 1, ± 2, …………

θ = 2nπ ± $$\frac{π}{3}$$ + $$\frac{π}{6}$$, where  n = 0, ± 1, ± 2, …………

Either, θ = 2nπ + $$\frac{π}{3}$$ + $$\frac{π}{6}$$ (4n + 1)$$\frac{π}{2}$$  ………..(1) or, θ = 2nπ - $$\frac{π}{3}$$ + $$\frac{π}{6}$$ = 2nπ - $$\frac{π}{6}$$  ………..(2)  Where 0, ± 1, ± 2, …………

Now, putting n = 0 in equation (1) we get, θ = $$\frac{π}{2}$$,

Putting n = 1 in equation (1) we get, θ = $$\frac{5π}{2}$$,

Putting  n = -1 in equation (1) we get, θ = - $$\frac{3π}{2}$$,

and putting  n = 0  in equation (2) we get, θ = - $$\frac{π}{6}$$

Putting n = 1 in equation (2) we get, θ = $$\frac{11π}{6}$$

Putting n = -1 in equation (2) we get, θ = - $$\frac{13π}{6}$$

Therefore,the  required solution of the trigonometric equation √3 cos θ + sin θ = 1 in -2π < θ < 2π are θ = $$\frac{π}{2}$$, - $$\frac{π}{6}$$, - $$\frac{3π}{2}$$, $$\frac{11π}{6}$$.