# Solve Trigonometry equations with cos x^0 + b sin x^0 = c

To
solve the equation in the form of a cos x

^{0}+ b sin x^{0}= c first, what must be done is to change the form of the equation a a cos x^{0}+ b sin x^{0}= c to be,
k cos (x - α)

^{0}= c
considering
– 1 ≤ cos (x - α) ≤ 1, so
that the equation can be solved must be fulfilled.

**Example 1.**

Determine
the set of completion equations cos x

^{0}– sin x^{0}= -1, for 0 ≤ x < 360.
cos x

^{0}– sin x^{0}= -1
√2 cos (x – 315)

^{0}= -1
cos (x – 315)

^{0}= -1/2 √2
x – 315 = 135 + n . 360 or x –
315 = - 135 + n . 360

x = 450 + n . 360 or x =
180 + n . 360

The
set of solutions is {90, 180}.

**Example 2**.

Determine
the set of completion equations – cos x

^{0}+ 2 sin x^{0}= 2, for 0≤ x ≤ 360.
Answer:

– cos x

^{0}+ 2 sin x^{0}= 2
√5 cos (x – 116.6)

^{0}= 2
cos (x – 116.6)

^{0}= 2 / √5
cos (x – 116.6)

^{0}= 0.894
x – 116.6 = 26.6 + n . 360 or x –
116.6 = -26.6 + n . 360

x = 143.2 + n . 360 or x =
90 + n . 360

The set of solutions
is {90, 143.2}.

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