Read the picture carefully the question is in!

3.sin² Ф + 13.cos² Ф = 7 ; 

13.cos² Ф can be written as : 3.cos²Ф + 10.cos² Ф

3.sin² Ф + 3.cos² Ф + 10.cos² Ф= 7 ; 
3(sin²Ф +cos²Ф) + 10 cos²Ф = 7; but (sin²Ф +cos²Ф) =1 , then:
3(1)  + 10.cos²Ф =7
10.cos²Ф = 4
cos²Ф = 4/10

cos(Ф) = √(4/10) = 0.632455

cos⁻¹(0.632455) = Ф = 0.8860 (in Radian)

Or in Degree , Ф = 50°76 (in Degree)

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